Question

$\int \frac{\sqrt{x}-\sqrt{a}}{\sqrt{x+a}}dx$ is equal to

• A. $\sqrt{x^2+ax}-\sqrt{ax+a^2}-acos{h}^{-1}\left(\sqrt{\frac{x+a}{a}}\right)+c$
• B. $\sqrt{x^2+ax}+\sqrt{ax+a^2}-a{\cosh }^{-1}\left(\sqrt{\frac{x+a}{a}}\right)+c$
• C. $\sqrt{x^2+ax}-2\sqrt{ax+a^2}+a{\cosh }^{-1}\left(\sqrt{\frac{x+a}{a}}\right)+c$
• D. $\sqrt{x^2-ax}+2\sqrt{ax-a^2}+a{\cosh }^{-1}\left(\sqrt{\frac{x+a}{a}}\right)+c$

Answer 1

The correct option is A $\sqrt{x^2+ax}-\sqrt{ax+a^2}-acos{h}^{-1}\left(\sqrt{\frac{x+a}{a}}\right)+c$

Let $I=\int \frac{\sqrt{x}-\sqrt{a}}{\sqrt{x+a}}dx$
Put $u=\sqrt{x}\Rightarrow du=\frac{1}{2\sqrt{x}}dx$
$I=2\int \frac{u(u-\sqrt{a})}{\sqrt{a+u^2}}du$
Put $u=\sqrt{a}\tan t\Rightarrow du=\sqrt{a}{\sec }^{2}tdt$
$\therefore I=2\sqrt{a}\int \tan t\sec t(\sqrt{a}\tan t-\sqrt{a})dt=2\sqrt{a}\int (\sqrt{a}{\tan }^{2}t\sec t-\sqrt{a}\tan t\sec t)dt=2a\int {\tan }^{2}t\sec tdt-2a\int \tan t\sec tdt=2a\int \sec t({\sec }^{2}t-1)dt-2a\int \tan t\sec tdt=2a\int {\sec }^{3}tdt-2a\int \sec tdt-2a\int \tan t\sec tdt$
Using reduction formula
$\int {\sec }^{m}xdx=\frac{\sin x{\sec }^{m-1}x}{m-1}+\frac{m-2}{m-1}\int {\sec }^{m-2}xdx$
$\therefore I=a\tan t\sec t-a\int \sec tdt-2a\int \tan t\sec tdt=-a\log (\tan t+\sec t)+a\tan t\sec t-2a\sec t$
$=\sqrt{x^2+ax}-\sqrt{ax+a^2}-a{\cosh }^{-1}\left(\sqrt{\frac{x+a}{a}}\right)+c$