Integration by parts: 3x^2(arctan x) dx.

I've shown my work here: http://i.imgur.com/LfA52.jpg

Where am I going wrong? Walk me through it step by step?

Question

Integration by parts:  3x^2(arctan x) dx.

I've shown my work here: http://i.imgur.com/LfA52.jpg

Where am I going wrong?  Walk me through it step by step?

anonymous · Answer

the problem is in your your first u, v substitution:

$\large dv=tan^{-1}xdx \rightarrow v=\frac{1}{x^2+1} $  is incorrect because 

$\large \frac{d}{dx}tan^{-1}(x)=\frac{1}{x^2+1} $.  what u need to do is integrate $\large  tan^{-1}x $ to get v.

anonymous · Answer

HELLO 94...
Change ur choice:
Let $$dv = 3x^2 ==> v = x^3 $$

anonymous · Answer

Aaand of c.  $$u(x) = \arctan (x) (=  \tan^{-1)}$$

anonymous · Answer

Then $$du = \frac{1}{1 + x^2}$$

anonymous · Answer

Aaand now, using integration by parts u will have 2 terms - first in which integral is already absent (difference of values) and the second will be$$-\int\limits \frac{x^3}{1 + x^2} = - \int\limits (x - \frac{x}{1+x^2}) = - \int\limits x + \int\limits \frac{x}{1+x^2} =-0.5x^{2} + \int\limits 0.5*d \ln(1+x^2)dx$$

anonymous · Answer

Hello @987456

anonymous · Answer

http://www.wolframalpha.com/input/?i=integrate+3x^2%28arctan+x%29 and click on show steps