Question

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