Question

Solve $\int \frac{x}{\sqrt{1-2x}}dx$

Answer 1

We have,

$I=\int \frac{x}{\sqrt{1-2x}}dx$

Let $t=1-2x$

$\frac{dt}{dx}=0-2$

$-\frac{dt}{2}=dx$

Therefore,

$I=-\frac{1}{2}\int \frac{(t-1)}{2\sqrt{t}}dt$

$I=\frac{1}{4}\int \frac{(1-t)}{\sqrt{t}}dt$

$I=\frac{1}{4}\int \frac{1}{\sqrt{t}}dt-\frac{1}{4}\int \sqrt{t}dt$

$I=\frac{1}{4}\left(2\sqrt{t}\right)-\frac{1}{4}\frac{{\left(t\right)}^{\frac{3}{2}}}{\frac{3}{2}}+C$

$I=\frac{1}{2}\left(\sqrt{t}\right)-\frac{1}{6}{\left(t\right)}^{\frac{3}{2}}+C$

On putting the value of $t$, we get

$I=\frac{1}{2}\left(\sqrt{1-2x}\right)-\frac{1}{6}{(1-2x)}^{\frac{3}{2}}+C$

Hence, this is the answer.