# How do you integrate (8x)/(4x^2+1)?

Aug 10, 2015

$\int \frac{8 x}{4 {x}^{2} + 1} \setminus \mathrm{dx} = \ln \left(4 {x}^{2} + 1\right) + C$

#### Explanation:

Do a substitution: let $u = 4 {x}^{2} + 1$ so $\mathrm{du} = 8 x \setminus \mathrm{dx}$. Therefore,

$\int \frac{8 x}{4 {x}^{2} + 1} \setminus \mathrm{dx} = \int \frac{1}{u} \setminus \mathrm{du} = \ln | u | + C$

$= \ln | 4 {x}^{2} + 1 | + C$.

Since $4 {x}^{2} + 1 \ge q 1 > 0$ for all $x \in \mathbb{R}$, we can get rid of the absolute value signs and write:

$\int \frac{8 x}{4 {x}^{2} + 1} \setminus \mathrm{dx} = \ln \left(4 {x}^{2} + 1\right) + C$