Question

If $\int \frac{x+1}{\sqrt{2x-1}}dx=f\left(x\right)\sqrt{2x-1}+C$, where $C$ is a constant of integration, then $f\left(x\right)$ is equal to :

• A. $\frac{1}{3}(x+1)$
• B. $\frac{2}{3}(x+2)$
• C. $\frac{2}{3}(x-4)$
• D. $\frac{1}{3}(x+4)$

Answer 1

The correct option is D $\frac{1}{3}(x+4)$

$I=\int \frac{x+1}{\sqrt{2x-1}}dx$
Let $2x-1=t^2$
$\Rightarrow dx=tdt$
$I=\int \frac{\frac{t^2+1}{2}+1}{t}\cdot tdt$
$=\frac{1}{2}\int (t^2+3)dt$

$=\frac{1}{2}[\frac{t^3}{3}+3t]+C$

Put $t=(2x-1{)}^{1/2}$
$I=\frac{1}{2}(2x-1{)}^{1/2}[\frac{2x-1}{3}+3]+C$
$=\frac{1}{2}(2x-1{)}^{1/2}\left[\frac{2x+8}{3}\right]+C$

Hence, $f\left(x\right)=\frac{x+4}{3}$