If $\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=\mathrm{f}(x) \sqrt{2 x-1}+\mathrm{c}$, (where c is a…
If $\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=\mathrm{f}(x) \sqrt{2 x-1}+\mathrm{c}$, (where c is a constant of integration), then $\mathrm{f}(x)$ is equal to
$\frac{1}{3}(x+1)$
$\frac{1}{3}(x+4)$
$\frac{2}{3}(x+2)$
$\frac{2}{3}(x-4)$
Solution
Let $\mathrm{I}=\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x$
Put $2 x-1=\mathrm{t}^2 \Rightarrow \mathrm{x}+1=\frac{\mathrm{t}^2+3}{2}$
$\begin{aligned}
\Rightarrow & d x=t \mathrm{dt} \\
\therefore \quad & =\int \frac{\left(\frac{\mathrm{t}^2+3}{2}\right) \mathrm{tdt}}{\mathrm{t}} \\
& =\frac{\mathrm{t}^3}{6}+\frac{3}{2} \mathrm{t}+\mathrm{c} \\
& =\frac{\mathrm{t}}{2}\left(\frac{\mathrm{t}^2+9}{3}\right)+\mathrm{c} \\
& =\frac{\sqrt{2 x-1}}{2}\left(\frac{2 x-1+9}{3}\right)+\mathrm{c} \\
& =\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+\mathrm{c}
\end{aligned}$
Comparing with $\mathrm{f}(x) \sqrt{2 x-1}+\mathrm{c}$, we get
$\mathrm{f}(x)=\frac{x+4}{3}$