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

Answer 1

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

$\sqrt{2x-1}=t⟹2x-1=t^2⟹2dx=2t.dt$
$\int \frac{x+1}{\sqrt{2x-1}}dx=\int \frac{\frac{t^2+1}{2}}{t}tdt=\int \frac{t^2+3}{2}$
$=\frac{1}{2}(\frac{t^3}{3}+3t)=\frac{t}{6}(t^2+9)+c$
$=\sqrt{2x-1}\left(\frac{2x-1+9}{6}\right)+c=\sqrt{2x-1}\left(\frac{x+4}{3}\right)+c$
$⟹f\left(x\right)=\frac{x+4}{3}$