... A good day to you Newton, If there were a Nobel prize for presenting, I would nominate you for it (lol) ... regarding converting x + 1 to 2x - 1, my way of thinking: x + 1 = 1/2*(2x + 2) = 1/2*(2x - 1 + 3), one can immediately see that factor 1/2 can be placed outside the integral without too much fuss ... I always jokingly call this "engineering the problem", and I often notice among young people that problems arise, even with these types of simple arithmetic actions, I on the other hand, had a strict arithmetic/maths upbringing, from which I'm still reaping the benefits! Friend Newton, I enjoyed your presentation, and thank you again for all your math efforts ... Stay well my friend, Jan-W
Nice! ....A more general way of reconstructing any intergrals of form "(px+q)/(ax^2+bx+c)" is px + q = A (d(ax 2 + bx + c)/dx) + B , where A and B can be obtained by comparing the coefficient.
Interesting manipulation to use the inverse tangent integral. I feel like it's simpler to use the int(1/(u^2 + a^2)) = 1/a arctan (u/a) formula. It would have cut out some steps at the end. Although, I can still appreciate the manipulation tricks used here.
This man is the savior.
Jesus is the Savior.
I appreciate the appreciation
... A good day to you Newton, If there were a Nobel prize for presenting, I would nominate you for it (lol) ... regarding converting x + 1 to 2x - 1, my way of thinking: x + 1 = 1/2*(2x + 2) = 1/2*(2x - 1 + 3), one can immediately see that factor 1/2 can be placed outside the integral without too much fuss ... I always jokingly call this "engineering the problem", and I often notice among young people that problems arise, even with these types of simple arithmetic actions, I on the other hand, had a strict arithmetic/maths upbringing, from which I'm still reaping the benefits! Friend Newton, I enjoyed your presentation, and thank you again for all your math efforts ... Stay well my friend, Jan-W
Sensational. You made it easy.
Nice! ....A more general way of reconstructing any intergrals of form "(px+q)/(ax^2+bx+c)" is px + q = A (d(ax 2 + bx + c)/dx) + B , where A and B can be obtained by comparing the coefficient.
Well done young man
More please. Thank you
Interesting manipulation to use the inverse tangent integral. I feel like it's simpler to use the int(1/(u^2 + a^2)) = 1/a arctan (u/a) formula. It would have cut out some steps at the end. Although, I can still appreciate the manipulation tricks used here.
I prefer the way Newton did it
Good job, Newton
I agree with you jason
You may also solve it without trig substitution by writing numerator of integrand 1 as 1+x^2-x^2 and then by separating the fractions
Nice vid
That was just perfect👏👏👏👏
❤