Mathematics - Indefinite Integration Question with Solution | TestHub

Let $I=\int \frac{x+1}{\sqrt{2x-1}}dx$ Put $\sqrt{2x-1}=t$ $\begin{array}{l}\therefore 2x-1=t^2\Rightarrow dx=tdt\\ I=\int \frac{\left(t^2+3\right)}{2}dt=\frac{t^3}{6}+\frac{3t}{2}+C\\ =\frac{(2x-1{)}^{\frac{3}{2}}}{6}+\frac{3}{2}(2x-1{)}^{\frac{1}{2}}+C\\ =\sqrt{2x-1}\left(\frac{x+4}{3}\right)+C\\ =f(x)\cdot \sqrt{2x-1}+C\end{array}$ Hence, $f(x)=\frac{x+4}{3}$