Question

If $\int \frac{\sqrt{x}}{x+1}dx=A\sqrt{x}+B{\tan }^{-1}\sqrt{x}+C,$ then $|A|+|B|$ is equal to
(where $C$ is integration constant and $A$ and $B$ are fixed constants)

• A. 4
• B. 4.00
• C. 4.0

Answer 1

Answer: (A), (B), (C)

Let,
$I=\int \frac{\sqrt{x}}{x+1}dx$
Put $x=t^2\Rightarrow dx=2tdt$
$\Rightarrow I=2\int \frac{t^2}{t^2+1}dt=2\int \frac{t^2+1-1}{t^2+1}dt=2\int (1-\frac{1}{t^2+1})dt=2[t-{\tan }^{-1}t]+C=2\sqrt{x}-2{\tan }^{-1}\sqrt{x}+C$
on comparing we get,
$A=2,B=-2$
$\therefore |A|+|B|=4$