Just here to say thanks from Brazil, just passed my calculus 2 exam! You're a great teacher and it's nice that you're solving all this different integrals on your channel, usualy we can only find the methods and simple examples on other channels, its awesome that we can get so much more practice on all the extra examples you're solving! I also dig how you smile at the camera from time to time, we get to see your passion for math and it kinda makes more bearable to learn from it, i already finished all my calculus 2 exams but im still here watching you, and i kinda hated calculus with all my might before lol. Keep killing it! Peace!
@@diegojesusduenascollado7043 that's not possible as we need it to be x^2 and not 4x^2, even if you try to take u = 4x^2, remember that to go to U world you have to differentiate u and write DX in terms of du, on doing that it'll change how the integral looks.
BPRP your channel is growing so much!!! Congrats! Also, I ended with an A in Differential Equations last semester thanks to you! Again thanks so much. This fall I'm taking linear algebra :)
you can solve by u sub if u sub in u = 2x+1 which yields ( after inserting x= (u-1)/2 and (2x+1) = u and cancelling everything and taking out the constants) 1/4e integral of ((ue^u-e^u)/u^2) du = integral of (e^u/u) - integral of (e^u/u^2) which is EI(u) - (EI(u)-e^u/u + c) = e^(2x+1)/e*4(1+2x) + c = e^2x/4(1+2x) + c
At 0:35 you say it can't be done by u substitution, but u = 1+2x works too and I found it simpler that way. It doesn't avoid having to do integral par parts though so it doesn't really save much time in the end.
Integration by partial fraction, dU is -ve and as well Integral V; so the product is positive. But there remains the negative sign from the formulation to take effect. Check at 6:48.
I have a tendency to "open math problems" like this. When I solve them already, I go online, look for other better way for them to learn extra procedure. In this particular problem, this remarkable professor and I have the same way. And I'm quite sure that is the best way for this problem. :)
Alternate method: The derivative of f(x)*(e^g(x)) = (e^g(x))*(g'(x)*f(x)+f'(x)). In this case g(x) = 2x, Hence we need to find f(x) such that 2f(x) + f'(x) = x/(1+2x)^2. By observation, f(x) should be of the form m/(1+2x). Putting this in the equation gives m=1/4. Hence we get the integral as (e^2x)/4(1+2x) + C
My man, thank you. I could not get past this problem in my homework. I had so many aha moments watching this video, and really every one of your videos that I've watched.
Saw 2x in a couple places, along with the single x up top, which can easily be made into 2x by multiplying by 2/2... I let u = 2x => du = 2dx . Subbing & factoring out 1/4, gives a cleaner: (1/4) integral[u e^u (1+u)^(-2) du]. The rest is easy.
You can do u-subs to u=1+2x too and separate the substituted equation into a sum of two integrals. Then, u have: I = 1/4e*[int(e^u/u)du - int(e^u/u^2)du**] So, if u apply integration by parts to first integral, taking e^u as dv and 1/u as w u have: int(e^u/u)du = e^u/u - int(-e^u/u^2)du int(e^u/u)du = e^u/u + int(e^u/u^2)du** Now, u can cancel (**). So: I=(1/(4e))(e^u/u)=e^(u-1)/4u = e^(2x)/(4(2x+1)) I think it is the first "blackpenredpen's integral" that i pwned alone, im very proud of that :)) thank you, your channel is awesome!
Good job! Linear substitutions simplification should have been obvious. There’s need to pull out the 1/e though. e^(u-1) is its own derivative and antiderivative, and the backsub is going to be happier when you’re done, too.
i prefer chalkboard over whiteboard, you can do many crazy things like drawing a circles with protractor and dotted lines with chalkboard, you cant do anything with whiteboard. whitechalkredchalk>blackpenredpen
I just stumbled across this video, and when you mentioned that partial fraction decomposition wasn't an option due to the exponential function in the numerator, it got me thinking. So I split the integrand into rational and non-rational parts, performed partial fraction decomposition on the rational part, and then redistributed the non-rational part, e^2x. It results in the same solution you arrived at, but the math is quite interesting because the exponential integral Ei(x) crops up twice in the expressions (Ei(2x+1)/4e) and -(Ei(2x+1)/4e), and so the terms containing the special function Ei(x) end up cancelling, and you're left with e^(2x)/4(2x+1) +C. Don't know if this was already mentioned, as I didn't read through several hundred comments LOL... Love the channel btw 👍
If you substitute u=2x+1, the integral reduces a lot, to the point where it breaks into two integrals. If we then use LIATE on the first integral, it eliminates the second integral, making life a LOT easier...
"product and quotient rule are the same thing" reminds me of 12th grade. amths teacher got pregnant we just had done product rule new maths teacher thought we were done with the chapter and in the exam was a shitload of fractions we had to derivate. man can you imagine the pain we were in?^^ i mean i for whatever reason remembered middleschool maths. and thouht i can just write the fraction u/v as u*v^-1 and differentiate this. man i was glad it only had simple stuff in the denominator like x^4. everything would have been physical pain without knowing quotient rule^^
you can use your trick to derive the quotient rule from the product rule if you have more than one quotient to differentiate. I had to do that when I suddenly had a memory failure during an exam.
i know nealry got hit for that from other guys in my semester. first semester chemistry. our prof asked who never heard of the quotient rule^^ some raised thweir hands. "hwo did you derive quotients then?" i said product rule and that guy "that would be a nice task for the exam thanks" we all fucked up^^ please derive the quotient rule from the product rule. i could do it in school but i could not do it correctly in theory in university man that sucked^^
Im a first year chem student (i have to do maths for 2 years tho but i love it) this seems like theperfect problem for someone my level chain rule product rule integration by parts and by substitution all used. Its perfect. And tge speed you teach at is perfect for me personally because im pretty comfterble with this. But man this is a perfect vid.
Hey plz tell me if you find this method useful :) I am solved it by using formula Integration { e^x [ f(x) +f'(x) ] .dx } = e^x [f(x)] +c 1. substitute 2x=u & simplify to get NOTE: There is 1/4 outside integration e^u [ u/ ( 1+u )^2] 2. add & subtract 1 in numerator to get it in this form e^u{ [1/(1+u)] + [ -1 / (1+u)^2] } 3.now comparing with formula we get f(x) as 1/(1+u) Therefore ans is obtained just by placing values in places of e^x[f(x)] +c
put 2x=t then you end up with [te^t/4(1+t)^2].... now we have e^t[(1/(1+t) - 1/(1+t)^2]/4... This takes the form of integral e^x[f(x)+f '(x)] = e^x[f(x)]...hence the answer [e^2x/(1+2x)]/4
I have a small query, at 5:50 when you write the first part of the answer, why is it negative? From the D column, the expression is positive and in the I column is the expression not also positive (because the negative sign in the expression cancels with the negative sign at the start of the second row)?
subs 2x=u and u get an integral in yhe the form of e^t(f(t)+f'(t)) by adding and subtracting 1 so that is just thr reverse product rule so answer is e^tf(t)
What criteria did you use to stop the derivation and integration of the two functions? Everything else made sense but I'm not sure exactly when to stop the integrating and derivating. Great videos btw
I solve this by factoring the numerator. In the integral, we see the denominator is in the from of “post-quotient rule” differentiation. I put A as the numerator of pre-differentiation, or the numerator of the resulting integral, then apply the quotient rule. Then we have: x.e^2x = A’(1+2x) - 2A Then, A’ -2A =0, and 2xA’ = xe^2x Then we will have A’ = [x(e^2x)]/2 and A = [x(e^2x)]/4 This is just for fun and fast for the math of multiplication question. I know the step and work it “backward”, but put it in proper explanation..kinda no😅
So, sorry for my obtuseness; but I'm confused about the technique that he is using here. I can understand the math in its' parts but what is he doing? Integration by parts? What method is he mentioning at 1:36 ? EDIT: NEVER MIND. I see that he is using integration by parts; he is just presenting it in a way that I was not accustomed to see it!
Sir. Please help me with regards this situation: integral of (-ln x /(1+e^2x)) dx at lower boundary of 0 to the upper boundary of 1. Please investigate this also if it is divergent or convergent. But I really do, it is convergent. But it is really tricky to solve because it cancels out. Thank you.
Well you’ve made an interesting linear differential equation here 😂 y’ + y[-2/(1+2x)] = xe^2x / (1+2x). I was trying to reverse engineer quotient rule and realized it was a differential equation. After finding e^(p(x))*q(x) I realized it was the same 🤣 Still, guessing it from this form is much easier. I’m doing this after the fact but it seems rather reasonable bc of the squared denom from Quotient rule (v = 2x+1, u=?): u’(2x+1) - 2u = xe^2x We need x multiplied by e^2x, so y definitely has e^2x. We don’t know the constant in front tho. Try for plugging in that, we get 4xe^x + 2e^2x - 2e^2x. Those two terms will always cancel out, so just divide by 4 and we get our answer. u/v = (e^2x /4)/(2x+1)
Hi. For these types of integrals you need to get the square root denominator term in one of these forms: √( a^2 + b^2 ) or √( a^2 - b^2 ) In your case (x^2+x+1)/√(2x^2-x+2) = (1/√2)( x^2 + x + 1 ) / √( ( x - 1/4)^2 + (15/16) ) If you require more help from here then please let me know.
The integration by parts formula is int u dv = uv - int v du, no? So, what happened to the minus before the int v du? I solved the question using substitution and then LIATE finding your answer but with a negative, viz, -e^(2x)/4(1+2x)^2 + C. Great videos btw!
Hi, before all thank u from the heart, but I have a question could we get the integral by the equation in the top with a negative exponent? in the first step. Because I just solve it by using this way and I had a different answer :(
"And the integral of this is...782 dollars!!!!!" - youtube ads have great timing
where to put the ads is chosen by the creators
@@ToonZIndia sometimes RUclips ignores the timings put in by creators and ads play at random moments
I think the category should be:
entertainment...
lol yea!
Just here to say thanks from Brazil, just passed my calculus 2 exam! You're a great teacher and it's nice that you're solving all this different integrals on your channel, usualy we can only find the methods and simple examples on other channels, its awesome that we can get so much more practice on all the extra examples you're solving! I also dig how you smile at the camera from time to time, we get to see your passion for math and it kinda makes more bearable to learn from it, i already finished all my calculus 2 exams but im still here watching you, and i kinda hated calculus with all my might before lol. Keep killing it! Peace!
Hi Miguel,
Thank you for your nice comment! I am glad that my videos have been help you! Keep up with you good work in your classes!
bprp
realy sympathetic teacher.. Thank you! Greetings from Germany
You're welcome!!
@@blackpenredpen Hi, isn't it possible that the second part of the integration can be integrated as arctg(2x)??? (1/1+(2x)²)
@@diegojesusduenascollado7043 that's not possible as we need it to be x^2 and not 4x^2, even if you try to take u = 4x^2, remember that to go to U world you have to differentiate u and write DX in terms of du, on doing that it'll change how the integral looks.
This dude is so good at teaching this stuff. My teacher is in no way a bad teacher, just seeing how someone else does it is so helpful.
*whitechalkredchalk
Rip pens
Yaayyy...
BPRP your channel is growing so much!!! Congrats! Also, I ended with an A in Differential Equations last semester thanks to you! Again thanks so much. This fall I'm taking linear algebra :)
zZwag that's great! Keep up with the good work!
zZwag I tried for like a minute to get that fly in ur pfp off of my screen lmao
you can solve by u sub if u sub in u = 2x+1 which yields ( after inserting x= (u-1)/2 and (2x+1) = u and cancelling everything and taking out the constants) 1/4e integral of ((ue^u-e^u)/u^2) du = integral of (e^u/u) - integral of (e^u/u^2) which is EI(u) - (EI(u)-e^u/u + c) = e^(2x+1)/e*4(1+2x) + c = e^2x/4(1+2x) + c
I love how the chalk colours match your sweater.
At 0:35 you say it can't be done by u substitution, but u = 1+2x works too and I found it simpler that way. It doesn't avoid having to do integral par parts though so it doesn't really save much time in the end.
Yes then it'd become in the form e^x(f(x)+f'(x)) which is e^x. f(x)
+C
Integration by partial fraction, dU is -ve and as well Integral V; so the product is positive. But there remains the negative sign from the formulation to take effect. Check at 6:48.
I have a tendency to "open math problems" like this.
When I solve them already, I go online, look for other better way for them to learn extra procedure.
In this particular problem, this remarkable professor and I have the same way. And I'm quite sure that is the best way for this problem. :)
Alternate method: The derivative of f(x)*(e^g(x)) = (e^g(x))*(g'(x)*f(x)+f'(x)). In this case g(x) = 2x, Hence we need to find f(x) such that 2f(x) + f'(x) = x/(1+2x)^2. By observation, f(x) should be of the form m/(1+2x). Putting this in the equation gives m=1/4. Hence we get the integral as (e^2x)/4(1+2x) + C
My man, thank you. I could not get past this problem in my homework. I had so many aha moments watching this video, and really every one of your videos that I've watched.
Saw 2x in a couple places, along with the single x up top, which can easily be made into 2x by multiplying by 2/2... I let u = 2x => du = 2dx . Subbing & factoring out 1/4, gives a cleaner:
(1/4) integral[u e^u (1+u)^(-2) du].
The rest is easy.
You can do u-subs to u=1+2x too and separate the substituted equation into a sum of two integrals. Then, u have:
I = 1/4e*[int(e^u/u)du - int(e^u/u^2)du**]
So, if u apply integration by parts to first integral, taking e^u as dv and 1/u as w u have:
int(e^u/u)du = e^u/u - int(-e^u/u^2)du
int(e^u/u)du = e^u/u + int(e^u/u^2)du**
Now, u can cancel (**). So:
I=(1/(4e))(e^u/u)=e^(u-1)/4u = e^(2x)/(4(2x+1))
I think it is the first "blackpenredpen's integral" that i pwned alone, im very proud of that :)) thank you, your channel is awesome!
Good job! Linear substitutions simplification should have been obvious. There’s need to pull out the 1/e though. e^(u-1) is its own derivative and antiderivative, and the backsub is going to be happier when you’re done, too.
Please do some eliptical integrals, I need to refresh them a bit. =)
Francisco Arias )
We can take
e^2x/1+2x = t
Differentiating it will give us
4 xe^2x/(1+2x)^2 dx = dt
which is basically the question. Its shorter
i prefer chalkboard over whiteboard, you can do many crazy things like drawing a circles with protractor and dotted lines with chalkboard, you cant do anything with whiteboard. whitechalkredchalk>blackpenredpen
But the particles would get into my nostril
And hands will be shabby
Thank you so much for posting your videos. They have helped me tremendously as I'm sure they have many many others.
that integral is so beautifully written!
¡El mejor explicando la solución de Integrales!
thanks!!!
I just stumbled across this video, and when you mentioned that partial fraction decomposition wasn't an option due to the exponential function in the numerator, it got me thinking. So I split the integrand into rational and non-rational parts, performed partial fraction decomposition on the rational part, and then redistributed the non-rational part, e^2x. It results in the same solution you arrived at, but the math is quite interesting because the exponential integral Ei(x) crops up twice in the expressions (Ei(2x+1)/4e) and -(Ei(2x+1)/4e), and so the terms containing the special function Ei(x) end up cancelling, and you're left with e^(2x)/4(2x+1) +C. Don't know if this was already mentioned, as I didn't read through several hundred comments LOL...
Love the channel btw 👍
this man is a living legend. you are the best
great video,
in other matter, can you do fourier transform videos like you did with laplace transform?
If you substitute u=2x+1, the integral reduces a lot, to the point where it breaks into two integrals. If we then use LIATE on the first integral, it eliminates the second integral, making life a LOT easier...
Dear bprp, congratulations on the example, the solution is surprising and fun !
Thanks
Elegant solution for a messy integral.
"product and quotient rule are the same thing" reminds me of 12th grade. amths teacher got pregnant we just had done product rule new maths teacher thought we were done with the chapter and in the exam was a shitload of fractions we had to derivate. man can you imagine the pain we were in?^^ i mean i for whatever reason remembered middleschool maths. and thouht i can just write the fraction u/v as u*v^-1 and differentiate this. man i was glad it only had simple stuff in the denominator like x^4. everything would have been physical pain without knowing quotient rule^^
you can use your trick to derive the quotient rule from the product rule if you have more than one quotient to differentiate. I had to do that when I suddenly had a memory failure during an exam.
i know nealry got hit for that from other guys in my semester. first semester chemistry. our prof asked who never heard of the quotient rule^^ some raised thweir hands. "hwo did you derive quotients then?" i said product rule and that guy "that would be a nice task for the exam thanks" we all fucked up^^ please derive the quotient rule from the product rule. i could do it in school but i could not do it correctly in theory in university man that sucked^^
I don't bother remembering the quotient rule. Products with negative powers work just as well.
Integral of x^2/(x sin x + cos x)^2, LIATE also seems to not work here but integration by parts works
Your solving methods are heart touching.❤️❤️❤️
Complicated question, complicated answer.lots of love from India
Im a first year chem student (i have to do maths for 2 years tho but i love it) this seems like theperfect problem for someone my level chain rule product rule integration by parts and by substitution all used. Its perfect. And tge speed you teach at is perfect for me personally because im pretty comfterble with this. But man this is a perfect vid.
You are a legend Sir, thank you so much for making these lessons easy to learn and concise. Wish us all luck in our Calc II exams!
So glad he put the +C in the end!
blackpenredpen's little brother... whitechalkredchalk
Perfect explanation with deep details I appreciate your work.
Hey plz tell me if you find this method useful :)
I am solved it by using formula
Integration { e^x [ f(x) +f'(x) ] .dx } = e^x [f(x)] +c
1. substitute 2x=u & simplify to get NOTE: There is 1/4 outside integration
e^u [ u/ ( 1+u )^2]
2. add & subtract 1 in numerator to get it in this form
e^u{ [1/(1+u)] + [ -1 / (1+u)^2] }
3.now comparing with formula we get f(x) as 1/(1+u)
Therefore ans is obtained just by placing values in places of e^x[f(x)] +c
This is visual asmr
fascinating. the way you flow with your maths. doing it with love. I'm inspired.
put 2x=t then you end up with [te^t/4(1+t)^2].... now we have e^t[(1/(1+t) - 1/(1+t)^2]/4... This takes the form of integral e^x[f(x)+f '(x)] = e^x[f(x)]...hence the answer [e^2x/(1+2x)]/4
What a great teacher you are, thanks for your uploads!
Thank you Art!
This integral came in my exam damn it!
I have a small query, at 5:50 when you write the first part of the answer, why is it negative? From the D column, the expression is positive and in the I column is the expression not also positive (because the negative sign in the expression cancels with the negative sign at the start of the second row)?
I had a video here explaining what to do in general
ruclips.net/video/2I-_SV8cwsw/видео.html
Hey, I didnt see his lecture, I just saw his cute handsome face
White Chalk Red Chalk is my favorite BPRP Variant.
I thought of 3 methods myself: this one, integrating x/(2x+1), and expanding the integrand.
subs 2x=u and u get an integral in yhe the form of e^t(f(t)+f'(t)) by adding and subtracting 1 so that is just thr reverse product rule so answer is e^tf(t)
1/(1+2x)^2 can be integrated as arctan(2x) with u=2x
Good. Your example helped me to solve another problem.
What criteria did you use to stop the derivation and integration of the two functions? Everything else made sense but I'm not sure exactly when to stop the integrating and derivating. Great videos btw
We integrate one time
You atop when it’s 1.easy/easier to integrate 2.you get your original function again
I solve this by factoring the numerator. In the integral, we see the denominator is in the from of “post-quotient rule” differentiation.
I put A as the numerator of pre-differentiation, or the numerator of the resulting integral, then apply the quotient rule.
Then we have: x.e^2x = A’(1+2x) - 2A
Then, A’ -2A =0, and 2xA’ = xe^2x
Then we will have A’ = [x(e^2x)]/2 and A = [x(e^2x)]/4
This is just for fun and fast for the math of multiplication question. I know the step and work it “backward”, but put it in proper explanation..kinda no😅
No necesitamos saber inglés para entender los números increíble explicación
I used u = 1 + 2x and it will lead to a Quotient Rule so just by checking you can get the solution.
Sir, u are awesome, pls can you upload more videos like these type
thank you so much blackpenredpen
So, sorry for my obtuseness; but I'm confused about the technique that he is using here. I can understand the math in its' parts but what is he doing? Integration by parts? What method is he mentioning at 1:36 ? EDIT: NEVER MIND. I see that he is using integration by parts; he is just presenting it in a way that I was not accustomed to see it!
Thank you for making our life easier .
Sir. Please help me with regards this situation: integral of (-ln x /(1+e^2x)) dx at lower boundary of 0 to the upper boundary of 1. Please investigate this also if it is divergent or convergent. But I really do, it is convergent. But it is really tricky to solve because it cancels out. Thank you.
*White Chalk, Red Chalk*
Thank you teacher
This problem is in the Stewart Calculus book
yea
A lot of the example problems he features are.
I think I saw your previous way because I did it by recognizing that it looked like a quotient so I did it by showing the quotient
Why doesn't the minus sign at the beginning of the second row cancel with the -1 in the numerator of the integrated fraction?
You saved my life 😭❤
Different way of thinking: Since (x e^2x/(2x+1))' = e^2x - 2x e^2x/(2x+1)^2, ∫x e^2x/(2x+1)^2 dx = (1/2)[(1/2) e^2x - x e^x/(2x+1)]
i see that supreme and that icy watch okayy professor
este men es un grande wn
Very niece explain sir👌👌👌
How did you know to stop after deriving and integrating once? Like don't we keep deriving and integrating?
What you said is really easy to understand .
Thx u so much.
At 3:40 he states that du = 2dx I dont get that part please help :(
He just differentiates (1+2x) from the 'u'
PLOT TWIST: HE HAS INFINITLY MANY BOARDS
I miss your class chow
Well you’ve made an interesting linear differential equation here 😂 y’ + y[-2/(1+2x)] = xe^2x / (1+2x). I was trying to reverse engineer quotient rule and realized it was a differential equation. After finding e^(p(x))*q(x) I realized it was the same 🤣
Still, guessing it from this form is much easier. I’m doing this after the fact but it seems rather reasonable bc of the squared denom from
Quotient rule (v = 2x+1, u=?):
u’(2x+1) - 2u = xe^2x
We need x multiplied by e^2x, so y definitely has e^2x. We don’t know the constant in front tho. Try for plugging in that, we get 4xe^x + 2e^2x - 2e^2x. Those two terms will always cancel out, so just divide by 4 and we get our answer.
u/v = (e^2x /4)/(2x+1)
my brain died on the second board
How can we get help or ask Questions, please help, if is possible
+blackpenredpen why the DI method notworks for (sec^2 x)(ln(cos(x))?
Its was the best video that i have wver watched 😊
Much love from 🇱🇰 ❤️
U didn't multiply integrated part by negative sign when u were taking product of cross elements.
crowd goes wild!
"Prada Rule"
Damn, I didn't know fashion was THAT strict!
U can solve this using-- integrate e^x( f(x) + f'(x))= e^x( f(x)) + c
Before BlackPenRedPen there was Chalk.
u are extremely fast and wonderfulll......
可以用前微後不動+後微前不動 d/dx(1/xe^x)
sir please solve this sums
integration of (x^2+x+1)/√(2x^2-x+2) dx. as soon as possible there is final exam .
Hi.
For these types of integrals you need to get the square root denominator term in one of these forms:
√( a^2 + b^2 )
or
√( a^2 - b^2 )
In your case (x^2+x+1)/√(2x^2-x+2) = (1/√2)( x^2 + x + 1 ) / √( ( x - 1/4)^2 + (15/16) )
If you require more help from here then please let me know.
OMG you are an awesome teacher I love you sir veryyyyyy much❤❤❤❤❤❤❤
How did the negative sign disappear?.. always remember to carry forward signs.
What do you mean he correctly did the sign
Substitute e^2x/(1+2x)
Use this trick you will get answer in 4 steps
The integration by parts formula is int u dv = uv - int v du, no? So, what happened to the minus before the int v du? I solved the question using substitution and then LIATE finding your answer but with a negative, viz, -e^(2x)/4(1+2x)^2 + C.
Great videos btw!
Damn is chalk satisfying to listen
question:
could you not just set u=e^2x
du/dx=e^2x
dx=1/e^2xdu
since u=e^2x, x=ln(u)/2
and 2x becomes ln(u)
to get the integral of ln(u)/2(1+ln(u))^2?
Thanks, I'm gonna have Calculus 2 exam in one week
You're perform maths in the best possible way :)
needs a link to the "D-I" method ?
Shit 🤦♂️🤦♂️💔💔 I didn't think of that... 1:40. I'll pause here
Hi, before all thank u from the heart, but I have a question could we get the integral by the equation in the top with a negative exponent? in the first step. Because I just solve it by using this way and I had a different answer :(
Instead of "black pen red pen" there is "white chalk red chalk".
Touche! Nicely Done Sir!!!