# How do you evaluate the integral int (xdx)/(5x^2-2)?

Feb 5, 2018

$\frac{1}{10} \ln | \left(5 {x}^{2} - 2\right) | + C$.

#### Explanation:

Prerequisite : $\int \left\{\frac{f ' \left(x\right)}{f} \left(x\right)\right\} \mathrm{dx} = \ln | f \left(x\right) | + c$.

This can be proved using subst. for $f \left(x\right)$.

Knowing taht, $\frac{d}{\mathrm{dx}} \left(5 {x}^{2} - 2\right) = 10 x$, we have,

$\int \frac{x \mathrm{dx}}{5 {x}^{2} - 2} = \frac{1}{10} \int \frac{10 x}{5 {x}^{2} - 2} \mathrm{dx}$,

$= \frac{1}{10} \ln | \left(5 {x}^{2} - 2\right) | + C$.

Feb 5, 2018

$\int \frac{x}{5 {x}^{2} - 2} \mathrm{dx} = \frac{1}{10} \ln | \left(5 {x}^{2} - 2\right) | + c$

#### Explanation:

we use the result

$\int \frac{f ' \left(x\right)}{f \left(x\right)} \mathrm{dx} = \ln | f \left(x\right) | + c$

$\int \frac{x}{5 {x}^{2} - 2} \mathrm{dx} - - \left(1\right)$

now $\frac{d}{\mathrm{dx}} \left(5 {x}^{2} - 2\right) = 10 x$

rewriting $\left(1\right)$

$\frac{1}{10} \int \frac{10 x}{5 {x}^{2} - 2} \mathrm{dx}$

we have

$\int \frac{x}{5 {x}^{2} - 2} \mathrm{dx} = \frac{1}{10} \ln | \left(5 {x}^{2} - 2\right) | + c$