Question

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

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

Answer 1

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

Let $I=\int \frac{(x-1)dx}{(x+1)\sqrt{(x^3+x^2+x)}}=\int \frac{(x^2-1)dx}{{(x+1)}^2\sqrt{x^3+x^2+x}}$
$=\int \frac{x^2(1-\frac{1}{x^2})dx}{(x^2+2x+1)\sqrt{x^3+x^2+x}}=\int \frac{x^2(1-\frac{1}{x^2})dx}{x(x+2+\frac{1}{x})x\sqrt{x+1+\frac{1}{x}}}$
Put $x+\frac{1}{x}=t\Rightarrow (1-\frac{1}{x^2})dx=dt$
Therefore
$I=\int \frac{dt}{(t+2)\sqrt{t+1}}$
Now put $1+t=u^2\Rightarrow dt=2udu$
Therefore
$I=\int \frac{2udu}{(u^2+1)\sqrt{u^2}}=2\int \frac{du}{u^2+1}$
$I=2{\tan }^{-1}u=2{\tan }^{-1}\left(\sqrt{1+t}\right)=2{\tan }^{-1}\sqrt{\frac{x^2+x+1}{x}}$
Hence, option 'A' is correct.