Question

If $\int \frac{(x-1)dx}{(x+1)\sqrt{x^3+x^2+x}}=f(x)+C$, then $f(1)=$

• A. $\frac{\pi }{4}$
• B. $\frac{2\pi }{5}$
• C. $\frac{2\pi }{3}$
• D. $\frac{5\pi }{6}$

Answer 1

Answer: (C)

Let $I=\int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}dx$
$=\int \frac{(x-1)(x+1)}{(x+1)(x+1)\sqrt{x^3+x^2+x}}dx$
$=\int \frac{x^2-1}{(x+1{)}^2\sqrt{x^3+x^2+x}}dx$
$=\int \frac{x^2-1}{\left(x^2+2x+1\right)\sqrt{x^3+x^2+x}}dx$
$=\int \frac{x^2\left(1-1/x^2\right)dx}{x\left(x+2+\frac{1}{x}\right)\cdot x\sqrt{x+1}+\frac{1}{x}}$
$=\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dx$
Put, $x+\frac{1}{x}+1=t^2$
$\Rightarrow \left(1-\frac{1}{x^2}\right)dx=2tdt$
$\therefore I=\int \frac{2tdt}{\left(t^2+1\right)t}=2\int \frac{dt}{t^2+1}=2{\tan }^{-1}t$
$=2{\tan }^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)+C$
$\therefore f(x)=2{\tan }^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)$
$\therefore f(1)=2{\tan }^{-1}\sqrt{3}=2\times \frac{\pi }{3}=\frac{2\pi }{3}$