Question

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

• A. $I=2{\tan }^{-1}\sqrt{\frac{x^2+x+1}{x}}+C$
• B. $I=2{\tan }^{-1}\sqrt{\frac{x^2+2x+1}{x}}+C$
• C. $I=2{\sec }^{-1}\sqrt{\frac{x^2+x+1}{x}}+C$
• D. $I=2{\sec }^{-1}\sqrt{\frac{x^2+2x+1}{x}}+C$

Answer 1

The correct option is A $I=2{\tan }^{-1}\sqrt{\frac{x^2+x+1}{x}}+C$

Let $I=\int \frac{(x-1)}{(x+1)\sqrt{x^3+x^2+x}}dx=\int \frac{(x^2-1)}{{(x+1)}^2\sqrt{x^3+x^2+x}}dx$
$=\int \frac{x^2(1-\frac{1}{x^2})}{(x^2+2x+1)\sqrt{x^3+x^2+x}}dx$
$=\int \frac{x^2(1-\frac{1}{x^2})}{x(x+2+\frac{1}{x}).x\sqrt{x+1+\frac{1}{x}}}dx$
Put $x+\frac{1}{x}=t$, $(1-\frac{1}{x^2})dx=dt$
$=\int \frac{dt}{(t+2)\sqrt{t+1}}$, which reduces to $\int \frac{dx}{P\sqrt{Q}}$.
Put $t+1=z^2$
$I=\int \frac{2zdz}{(z^2+1)\sqrt{z^2}}=2\int \frac{dz}{z^2+1}=2{\tan }^{-1}\left(z\right)+C$
$=2{\tan }^{-1}\left(\sqrt{t+1}\right)+C=2{\tan }^{-1}\sqrt{\frac{x^2+x+1}{x}}+C$