Question

Evaluate $\int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}dx$

• A. ${\tan }^{-1}\sqrt{(x+\frac{1}{x}+1)}+c$
• B. $2{\tan }^{-1}\sqrt{(x+\frac{1}{x}+1)}+c$
• C. ${\tan }^{-1}\sqrt{(x+\frac{1}{x}-1)}$
• D. ${\tan }^{-1}\sqrt{(x-\frac{1}{x}+1)}+c$

Answer 1

The correct option is B $2{\tan }^{-1}\sqrt{(x+\frac{1}{x}+1)}+c$

$I=\int \frac{(x-1)dx}{(x+1)\sqrt{x^3+x^2+x}}$
$=\int \frac{(x^2-1)}{(x^2+2x+1)\sqrt{x^3+x^2+x}}$
$=\int \frac{(1-\frac{1}{x^2})dx}{(x+\frac{1}{x}+2)\sqrt{x+\frac{1}{x}+1}}$
Substitute $x+\frac{1}{x}+1={\mu }^2,$
$I=\int \frac{2\mu du}{({\mu }^2+1)\mu }=2{\tan }^{-1}\mu +c$
$=2{\tan }^{-1}\sqrt{(x+\frac{1}{x}+1)}+c$