If $\int \frac{\mathrm{x}+1}{\sqrt{2 \mathrm{x}-1}} \mathrm{dx}=\mathrm{f}(\mathrm{x}) \sqrt{2 \mathrm{x}-1}+\mathrm{C}$, where $C$ is a
constant of integration, then $\mathrm{f}(\mathrm{x})$ is equal to :-
Correct Option: 1
$\sqrt{2 \mathrm{x}-1}=\mathrm{t} \Rightarrow 2 \mathrm{x}-1=\mathrm{t}^{2} \Rightarrow 2 \mathrm{dx}=2 \mathrm{t} \cdot \mathrm{dt}$
$\int \frac{x+1}{\sqrt{2 x-1}} d x=\int \frac{\frac{t^{2}+1}{2}+1}{t} t d t=\int \frac{t^{2}+3}{2} d t$
$=\frac{1}{2}\left(\frac{t^{3}}{3}+3 t\right)=\frac{t}{6}\left(t^{2}+9\right)+c$
$=\sqrt{2 x-1}\left(\frac{2 x-1+9}{6}\right)+c=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+c$
$\Rightarrow f(x)=\frac{x+4}{3}$
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