# What is the antiderivative of (2x)/(sqrt(5 + 4x))?

Aug 23, 2016

$I = \frac{1}{8} \left\{5 + 4 x - 10 \ln \sqrt{5 + 4 x}\right\} + C$, or,

$I = \frac{1}{4} \left(2 x - 5 \ln \sqrt{5 + 4 x}\right) + K$, where, $K = \frac{5}{8} + C$.

#### Explanation:

We take substn. $5 + 4 x = {t}^{2} , s o t \hat{,} 4 \mathrm{dx} = 2 t \mathrm{dt} , i . e . , \mathrm{dx} = \frac{1}{2} \mathrm{dt}$.

Also, $5 + 4 x = {t}^{2} \Rightarrow x = \frac{{t}^{2} - 5}{4}$. Therefore,

$I = \int \frac{2 x}{\sqrt{5 + 4 x}} \mathrm{dx} = \frac{1}{2} \int \frac{2 \frac{{t}^{2} - 5}{4}}{\sqrt{{t}^{2}}} \mathrm{dt} = \frac{1}{4} \int \frac{{t}^{2} - 5}{t} \mathrm{dt}$

$= \frac{1}{4} \int \left({t}^{2} / t - \frac{5}{t}\right) \mathrm{dt} = \frac{1}{4} \left({t}^{2} / 2 - 5 \ln | t |\right)$

$= \frac{1}{8} \left({t}^{2} - 10 \ln | t |\right)$. Therefore,

$I = \frac{1}{8} \left\{5 + 4 x - 10 \ln \sqrt{5 + 4 x}\right\} + C$

Readjusting the Constant of Integration, we may like,

$I = \frac{1}{4} \left(2 x - 5 \ln \sqrt{5 + 4 x}\right) + K$, where, $K = \frac{5}{8} + C$.

Enjoy Maths.!