Question

$\int \frac{x}{(\sqrt{x+a}+\sqrt{x+b})}dx$ is equal to

• A. $\frac{1}{a+b}[\frac{5}{2}{(x+a)}^{\frac{5}{2}}-\frac{3a}{2}{(x+a)}^{\frac{2}{3}}-\frac{5}{2}{(x+b)}^{\frac{2}{5}}+\frac{3}{2}b{(x+b)}^{\frac{2}{3}}]+C$
• B. $\frac{1}{a+b}[\frac{2}{5}{(x-a)}^{\frac{5}{2}}-\frac{2a}{3}{(x-a)}^{\frac{3}{2}}-\frac{2}{5}{(x-b)}^{\frac{5}{2}}+\frac{2}{3}b{(x-b)}^{\frac{3}{2}}]+C$
• C. $\frac{1}{a-b}[\frac{2}{5}{(x+a)}^{\frac{5}{2}}-\frac{2a}{3}{(x+a)}^{\frac{3}{2}}-\frac{2}{5}{(x+b)}^{\frac{5}{2}}+\frac{2}{3}b{(x+b)}^{\frac{3}{2}}]+C$
• D. none of these

Answer 1

The correct option is C $\frac{1}{a-b}[\frac{2}{5}{(x+a)}^{\frac{5}{2}}-\frac{2a}{3}{(x+a)}^{\frac{3}{2}}-\frac{2}{5}{(x+b)}^{\frac{5}{2}}+\frac{2}{3}b{(x+b)}^{\frac{3}{2}}]+C$

$I=\int \frac{x}{(\sqrt{x+a}+\sqrt{x+b})}dx$
$=\int \frac{x(\sqrt{x+a}-\sqrt{x+b})}{(x+a)-(x+b)}dx$
$=\frac{1}{a-b}\int \left(x\sqrt{x+a}-x\sqrt{x+b}\right)dx$
$=\frac{1}{a-b}\int \left\{\right(x+a-a)\sqrt{x+a}-(x+b-b\left)\sqrt{x+b}\right\}dx$
$=\frac{1}{a-b}\int \{{(x+a)}^{3/2}-a{(x+a)}^{1/2}-{(x+b)}^{3/2}+b{(x+b)}^{1/2}\}dx$
$=\frac{1}{a-b}[\frac{2}{5}{(x+a)}^{5/2}-\frac{2a}{3}{(x+a)}^{3/2}-\frac{2}{5}{(x+b)}^{5/2}+\frac{2}{3}b{(x+b)}^{3/2}]+C$