# How do you use Integration by Substitution to find intx/(x^2+1)dx?

Mar 16, 2018

$\int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \ln \left({x}^{2} + 1\right) + \text{c}$

#### Explanation:

We want of find $\int \frac{x}{{x}^{2} + 1} \mathrm{dx}$.

We make the natural substitution $u = {x}^{2} + 1$ and $\mathrm{du} = 2 x \mathrm{dx}$.

So

$\int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \int \frac{1}{u} \mathrm{du} = \frac{1}{2} \ln \left\mid u \right\mid = \frac{1}{2} \ln \left\mid {x}^{2} + 1 \right\mid + \text{c}$

We notice that ${x}^{2} + 1 > 0$ so we don't need to take the absolute value. So the final answer is

$\frac{1}{2} \ln \left({x}^{2} + 1\right) + \text{c}$

Mar 19, 2018

You don't. You simply write down the answer.

#### Explanation:

$\int \frac{x}{{x}^{2} + 1} \setminus \mathrm{dx}$
$= \left(\frac{1}{2}\right) \int \frac{2 x}{{x}^{2} + 1}$
The integral is of the form $\int \frac{f ' \left(x\right)}{f} \left(x\right) \setminus \mathrm{dx} = \ln | f \left(x\right) |$
So you can just write down the answer, dropping the $| |$, as ${x}^{2} + 1 > 0$:

$\frac{1}{2} \ln \left({x}^{2} + 1\right) + c$
or alternatively $\ln \sqrt{{x}^{2} + 1} + c$
With practice you can do this type in one step at sight!