Hi! I don't see where I forgot to write the 2ln(x)... could you tell me where? On the other hand, if I calculate the derivative of the solution I get the original question...: Derivative of (x^2/2)ln^2(x) - (x^2/2)ln(x) + x^2/4 + C = = (1/2)*Derivative of (x^2)ln^2(x) - (1/2)*Derivative of (x^2)ln(x) + (1/4)*Derivative of x^2 = = (1/2)*( (x^2)*2ln(x)*(1/x) + 2x*ln^2(x) ) - (1/2)*( (x^2)*(1/x) + 2x*ln(x) ) + (1/4)*2x = = (1/2)*( 2x*ln(x) + 2x*ln^2(x) ) - (1/2)*( x + 2x*ln(x) ) + (1/2)*x = = x*ln(x) + x*ln^2(x) - (1/2)*x - x*ln(x) + (1/2)*x = = x*ln^2(x)
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Keep up the good work brother! Greetings for Brazil! THANK U
Thanks, Luis! ;-D
Greetings for Brazil or from Brazil? Try improving English brother ....
Thanks, you have helped me a lot. This question is in the book of a Brazilian author called Guidorizze.
I'm glad it helped! 💪
Hello, you forgot to write "2 Ln(x)" in du you only wrote 1/x dx
therefore your answer is incorrect :(
Hi! I don't see where I forgot to write the 2ln(x)... could you tell me where?
On the other hand, if I calculate the derivative of the solution I get the original question...:
Derivative of (x^2/2)ln^2(x) - (x^2/2)ln(x) + x^2/4 + C =
= (1/2)*Derivative of (x^2)ln^2(x) - (1/2)*Derivative of (x^2)ln(x) + (1/4)*Derivative of x^2 =
= (1/2)*( (x^2)*2ln(x)*(1/x) + 2x*ln^2(x) ) - (1/2)*( (x^2)*(1/x) + 2x*ln(x) ) + (1/4)*2x =
= (1/2)*( 2x*ln(x) + 2x*ln^2(x) ) - (1/2)*( x + 2x*ln(x) ) + (1/2)*x =
= x*ln(x) + x*ln^2(x) - (1/2)*x - x*ln(x) + (1/2)*x =
= x*ln^2(x)
Caraca velho, essa é a questão H do GUIDORIZZI, muito boa! ;)
:-D
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If u say what you are writing and maybe explain it .it would be much better .
Thanks for all of your great videos
Thanks for your advice, Ahmed! I am thinking of improving it and one day I will! ;-D
Gracias
i hate maths, but thankyou for this
If you don't hate me it is ok for me 😜Good luck with integration, if you have questions, I'm here 😉