# How do you find the indefinite integral of int x/(x^2+1)?

Dec 4, 2016

$\int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \ln \sqrt{{x}^{2} + 1} + C$

#### Explanation:

With a little experience you may be able to see that the numerator is almost the derivative of the denominator, and we can use that:

$\int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \int \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$
$\therefore \int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \ln | {x}^{2} + 1 | + C$
$\therefore \int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \ln \left({x}^{2} + 1\right) + C$ (As ${x}^{2} + 1 > 0$)
$\therefore \int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \ln \sqrt{{x}^{2} + 1} + C$

If you can't spot that feature then we can use a substitution:

Let $u = {x}^{2} + 1$, Then $\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$
So, "separating the variables" we get :

$\int \ldots \mathrm{du} = \int \ldots 2 x \mathrm{dx} \implies \int \ldots x \mathrm{dx} = \frac{1}{2} \int \ldots \mathrm{du} =$

And so substituting into the original integral:

$\int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \int \frac{1}{u} \mathrm{du}$
$\therefore \int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \ln | u | + C$
$\therefore \int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \ln | {x}^{2} + 1 | + C$, as before