SOLUTION: it's actually about calculus but i couldn't find any parts about it.. i was solving integration of tan^-1(x) by parts : consider ∫tan^-1(x)x1 dx. so let u=tan^-1(x) and du

Algebra ->  Trigonometry-basics -> SOLUTION: it's actually about calculus but i couldn't find any parts about it.. i was solving integration of tan^-1(x) by parts : consider ∫tan^-1(x)x1 dx. so let u=tan^-1(x) and du      Log On


   

Question 939809: it's actually about calculus but i couldn't find any parts about it..
i was solving integration of tan^-1(x) by parts :
consider ∫tan^-1(x)x1 dx.
so let u=tan^-1(x) and du/dx=1/(1^2+x^2)
dv/dx=1, so v=x.
∫tan^-1(x)dx = xtan^-1(x)-∫(1/(1+x^2))(x)dx
= xtan^-1(x)-∫(x/(1+x^2)dx
= xtan^-1(x)-(1/2)∫(2x/(1+x^2))dx
= xtan^-1(x)-(1/2)tan^-1(x^2)+c.
but the answer is xtan^-1(x)-(1/2)ln(1+(x^2))+c.
Why???
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Note the integration by parts formula:

In this case, since and (you forgot x in the numerator)

Integrating using the substitution gives , as desired.