Stumped by an Integral

**adkinsjr** · December 7th 2009, 03:25 PM

I am trying to evaluate the following:

$\int\frac{2+x^2}{1+x^2}dx$

So far I have:

$\int\frac{2}{1+x^2}dx+\int\frac{x^2}{1+x^2}dx$

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

Obviously I'm stuck on the second term. Any ideas?

**adkinsjr** · December 7th 2009, 03:31 PM

$\int\frac{2+x^2}{1+x^2}dx=\frac{1}{2}\int\frac{1+u }{\sqrt{u}(1+u)}du=\frac{1}{2}\int\frac{1}{\sqrt{u }}du$

$u=x^2$

nevermind,

**Defunkt** · December 7th 2009, 03:32 PM

$\int\frac{2+x^2}{1+x^2}dx = \int\frac{1+x^2 + 1}{1+x^2}dx =$ $\int 1 +\frac{1}{1+x^2} dx$

And $\int\frac{1+x^2}{1+x^2}dx = \int 1dx$

**adkinsjr** · December 7th 2009, 03:46 PM

Yeah, that last one was a typo. I edited it. I meant for 2-x^2 to be in the numberator. thanks

**Defunkt** · December 8th 2009, 12:58 AM

Oops, I misunderstood your last post.

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