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 $f(x)$ 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

Answer: (A)

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