What's wrong with my integration?

I've used integration by parts:

$$\int x\log ({x^2} + 1)dx \to \frac{{{x^2}}}{2}\log ({x^2} + 1) + \int \frac{{2{x^3}}}{{2{x^2} + 2}}$$

Then the second integral I used polynomial division and got:

$$\frac{{{x^2}}}{2} - log(2 + {x^2} + c)$$

Sticking everything together to get:

$$\frac{{{x^2}}}{2}\log ({x^2} + 1) + \frac{{{x^2}}}{2} - log(2 + {x^2}) + c$$

Question

What's wrong with my integration?

I've used integration by parts:

$$\int x\log ({x^2} + 1)dx \to \frac{{{x^2}}}{2}\log ({x^2} + 1) + \int \frac{{2{x^3}}}{{2{x^2} + 2}}$$

Then the second integral I used polynomial division and got:

$$\frac{{{x^2}}}{2} - log(2 + {x^2} + c)$$

Sticking everything together to get:

$$\frac{{{x^2}}}{2}\log ({x^2} + 1) + \frac{{{x^2}}}{2} - log(2 + {x^2}) + c$$

alfie · Answer

wolfram doesn't seem to agree with me: http://www.wolframalpha.com/input/?i=integrate%20xlog%28x^2%2B1%29dx&t=mfftb01

alfie · Answer

The  + c wasn't supposed to be inside the parenthesis, equation typo >.<

amistre64 · Answer

x+3/(x^2+1) is simplier i think

amistre64 · Answer

x^3, not x+3 ... fat fingers

amistre64 · Answer

and should be a - integration x^3/.... i believe

amistre64 · Answer

$$\int x\log ({x^2} + 1)dx = \frac{{{x^2}}}{2}\log ({x^2} + 1) - \int \frac{x^2}{2}*\frac{2x}{x^2+1}dx $$

$$\int x\log ({x^2} + 1)dx = \frac{{{x^2}}}{2}\log ({x^2} + 1) - \int \frac{x^3}{x^2+1}dx $$

alfie · Answer

Yup, I failed at the second integration (I factored out and left the +2 there, silly me), sec, let me do it and type it down :)

slaaibak · Answer

It would have been easier on the eye if you first substituted x^2 + 1 = u

alfie · Answer

Okay, I did it, it's all right I failed at factoring out (ridiculous >.<)
Thank you amistre.

amistre64 · Answer

is that log natural or base 10?

amistre64 · Answer

youre welcome :)