# Math Help - int x/(x^2+1)

1. ## int x/(x^2+1)

hello!

I have to solve :
$\int \frac{x}{x^2+1}dx=\frac{1}{2}ln|x^2+1|+C$

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

if weknow that:
$\int u^{-1}du=\int \frac{1}{u}du$ and $u=x^2+1$

then what do we exactly do with 2xdx in the numerator? Dont we just have 2x too much???

many thanks!

2. ## Re: int x/(x^2+1)

It's because when you substitute \displaystyle \begin{align*} u = x^2 + 1 \end{align*}, that means \displaystyle \begin{align*} \frac{du}{dx} = 2x \end{align*}. So

\displaystyle \begin{align*} \int{\frac{x}{x^2 + 1}\,dx} &= \frac{1}{2}\int{ \frac{1}{x^2 + 1}\cdot 2x \, dx} \\ &= \frac{1}{2} \int{ \frac{1}{u} \, \frac{du}{dx} \, dx } \\ &= \frac{1}{2}\int{ \frac{1}{u}\,du} \end{align*}

which you can now integrate.

Thanks!