Question

Evaluate:

$\int \left(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\right)dx$

• A. $\frac{3}{2(a-b)}((x+a{)}^{3/2}-(x+b{)}^{3/2})+c$
• B. $\frac{2}{3(a-b)}((x+a{)}^{3/2}-(x+b{)}^{3/2})+c$
• C. $\frac{2}{3(a-b)}((x-a{)}^{3/2}-(x+b{)}^{3/2})+c$
• D. $\frac{2}{3(a-b)}((x-a{)}^{3/2}-(x-b{)}^{3/2})+c$

Answer 1

The correct option is B $\frac{2}{3(a-b)}((x+a{)}^{3/2}-(x+b{)}^{3/2})+c$

$\int \frac{1}{\sqrt{x+a}\sqrt{x+b}}dx$

$\int \frac{\sqrt{x+a}-\sqrt{x+b}}{x+a-x-b}dx$

$\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}dx$

$\frac{1}{a-b}\int \sqrt{x+a}dx-\frac{1}{a-b}\int \sqrt{x+b}dx$

$=\frac{1}{a-b}\frac{(x+a{)}^{3/2}}{3/2}-\frac{1}{a-b}\frac{(x+b{)}^{3/2}}{3/2}+c$

$=\frac{2}{3(a-b)}(x+a{)}^{3/2}-\frac{2}{3(a-b)}(x+b{)}^{3/2}+c$

$=\frac{2}{3(a-b)}\left[\right(x+a{)}^{3/2}-(x+b{)}^{3/2}]+c$