Question

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

• A. ${\tan }^{-1}(x+\frac{1}{x}+1)+C$
• B. ${\tan }^{-1}\sqrt{x+\frac{1}{x}+1}+C$
• C. $2{\tan }^{-1}\sqrt{x+\frac{1}{x}+1}+C$
• D. none of these

Answer 1

Answer: (C)

$2{\tan }^{-1}\sqrt{x+\frac{1}{x}+1}+C$

Let $I=\int \frac{(x-1)dx}{(x+1)\sqrt{x^3+x^2+x}}=\int \frac{(x^2-1)dx}{{(x+1)}^2\sqrt{x^3+x^2+x}}$
$=\int \frac{x^2(1-\frac{1}{x^2})dx}{(x^2+2x+1)\sqrt{x^3+x^2+x}}=\int \frac{x^2(1-\frac{1}{x^2})dx}{x(x+2+\frac{1}{x}).x.\sqrt{x+1+\frac{1}{x}}}$
Substitute $x+\frac{1}{x}=t\Rightarrow (1-\frac{1}{x^2})dx=dt$
Let $I=\int \frac{dt}{(t+2)\sqrt{t+1}}$
Substitute $1+t=z^2\Rightarrow dt=2zdz$
$I=\int \frac{2zdz}{(z^2+1)\sqrt{z^2}}=2\int \frac{dz}{z^2+1}=2{\tan }^{-1}\left(z\right)+c$
$=2{\tan }^{-1}\sqrt{1+t}+c=2{\tan }^{-1}\sqrt{\frac{x^2+x+1}{x}}+c$