Question

If $I=\int ta{n}^{-1}\sqrt{(\sqrt{x}-1)}dx=(u^2+1{)}^{2}ta{n}^{-1}u-\frac{A}{1863}{u}^3-u+C$ where $u=\sqrt{\sqrt{x}-1}$ then A is equal to.

• A. 321
• B. 312
• C. 316
• D. 318

Answer 1

The correct option is A 321

$I=\int {\tan }^{-1}\left(\sqrt{\sqrt{x}-1}\right)dx=x{\tan }^{-1}\left(\sqrt{\sqrt{x}-1}\right)-\frac{1}{4}\int \frac{1}{\sqrt{\sqrt{x}-1}}dx$
Let $I_1=-\frac{1}{4}\int \frac{1}{\sqrt{\sqrt{x}-1}}dx$
Put $t=\sqrt{x}\Rightarrow dx=\frac{1}{2\sqrt{x}}dx$
$I_1=-\frac{1}{2}\int \frac{t}{\sqrt{t-1}}dt$
Put $v=t-1\Rightarrow dv=dt$
$I_1=-\frac{1}{2}\int \frac{v+1}{\sqrt{v}}du=-\frac{1}{2}\int (\sqrt{v}+\frac{1}{\sqrt{v}})du=-\frac{v^{3/2}}{3}-\sqrt{v}+c=-\frac{{(t-1)}^{3/2}}{3}-\sqrt{t-1}+c=-\frac{{(\sqrt{x}-1)}^{3/2}}{3}-\sqrt{\sqrt{x}-1}+c$
Hence
$I=x{\tan }^{-1}\left(\sqrt{\sqrt{x}-1}\right)-\frac{{(\sqrt{x}-1)}^{3/2}}{3}-\sqrt{\sqrt{x}-1}+c={(u^2+1)}^2{\tan }^{-1}u-\frac{u^3}{3}-u+c$
Where $u=\sqrt{\sqrt{x}-1}$
Therefore $A=321$