# Integration of x/(2x+5)?

Apr 7, 2018

$I = \frac{1}{4} \left(\left(2 x + 5\right) - 5 \ln \left(\left\mid \left(2 x + 5\right) \right\mid\right)\right) + C$

#### Explanation:

We want to solve

$I = \int \frac{x}{2 x + 5} \mathrm{dx}$

Make a substitution color(blue)(u=2x+5=>du=2dx

$I = \frac{1}{2} \int \frac{x}{u} \mathrm{du}$

But color(blue)(u=2x+5=>x=(u-5)/2

$I = \frac{1}{2} \int \frac{\frac{u - 5}{2}}{u} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \int \frac{u - 5}{u} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \int 1 - \frac{5}{u} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \left(u - 5 \ln \left(\left\mid u \right\mid\right)\right) + C$

Substitute back $u = \left(2 x + 5\right)$

$I = \frac{1}{4} \left(\left(2 x + 5\right) - 5 \ln \left(\left\mid \left(2 x + 5\right) \right\mid\right)\right) + C$

Apr 7, 2018

$I = \int \frac{x}{2 x + 5} \mathrm{dx}$

Let $2 x + 5 = t$

Therefore, $\mathrm{dt} = 2 \mathrm{dx} \implies \mathrm{dx} = \frac{\mathrm{dt}}{2}$

Also, $x = \frac{t - 5}{2}$

$\implies I = \int \frac{t - 5}{2 \left(t\right)} \frac{\mathrm{dt}}{2}$

$\implies \frac{1}{4} \int \frac{t - 5}{t} \mathrm{dt}$

$\implies \frac{1}{4} \left[\int 1 \mathrm{dt} - 5 \int \frac{1}{t} \mathrm{dt}\right]$

$\implies \frac{1}{4} \left[t - 5 \log | t |\right] + c$

$\implies \frac{1}{4} \left(2 x + 5\right) - \frac{5}{4} \log | 2 x + 5 | + c$