Practicing Success
Practicing Success
CUET Preparation Today
$\int \frac{d x}{(2 x+1)(1+\sqrt{(2 x+1)}}$ is equal to : |
$\tan ^{-1} \frac{\sqrt{2 x+1}}{1+\sqrt{2 x+1}}+c$ $\log _{e} \frac{\sqrt{2 x+1}}{1+\sqrt{2 x+1}}+c$ $\log _{e}\left(\frac{1+\sqrt{2 x+1}}{\sqrt{2 x+1}}\right)+c$ $\tan ^{-1} \frac{1+\sqrt{2 x+1}}{\sqrt{2 x+1}}+c$ |
$\log _{e} \frac{\sqrt{2 x+1}}{1+\sqrt{2 x+1}}+c$ |
$I=\int \frac{d x}{(2 x+1)(1+\sqrt{(2 x+1)}}$ so let $y=\sqrt{2x+1}$ $dy=\frac{1}{\sqrt{2x+1}}dx$ $⇒I=\int\frac{dy}{y(1+y)}=\int\frac{1}{y}-\frac{1}{1+y}dy$ $=\log y-\log 1+y+C$ $=\log\frac{y}{1+y}+C=\log\frac{\sqrt{2x+1}}{1+\sqrt{2x+1}}+C$ |
