Question:
If $\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=f(x) \sqrt{2 x-1}+\mathrm{C}$, where $\mathrm{C}$ is a constant
of integration, then $f(x)$ is equal to:
Correct Option: , 3
Solution:
Let $I=\int \frac{x+1}{\sqrt{2 x-1}} d x$
Put $\sqrt{2 x-1}=t$
$\therefore \quad 2 x-1=t^{2} \Rightarrow d x=t d t$
$I=\int \frac{\left(t^{2}+3\right)}{2} d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C$
$=\frac{(2 x-1)^{\frac{3}{2}}}{6}+\frac{3}{2}(2 x-1)^{\frac{1}{2}}+C$
$=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+C$
$=f(x) \cdot \sqrt{2 x-1}+C$
Hence, $f(x)=\frac{x+4}{3}$