Question

The value of $\int \frac{x^4-1}{x^2{(x^4+x^2+1)}^{\frac{1}{2}}}dx$ is equal to

• A. $\sqrt{\frac{x^4+x^2+1}{x}}+C$
• B. $\sqrt{\frac{x^4+x^2+1}{x^3}}+C$
• C. $\sqrt{\frac{x^4+x^2+1}{x^2}}+C$
• D. None of these

Answer 1

The correct option is C $\sqrt{\frac{x^4+x^2+1}{x^2}}+C$

$I=\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx=\int \frac{(x^2-1)(x^2+1)}{x^3\sqrt{x^2+\frac{1}{x^2}+1}}dx=\int \frac{(x+\frac{1}{x})(1-\frac{1}{x^2})}{\sqrt{{(x+\frac{1}{x})}^2-1}}dx$
Putting ${(x+\frac{1}{x})}^2-1=t\Rightarrow 2(x+\frac{1}{x})(1-\frac{1}{x^2})dx=dt$
$I=\int \frac{dt}{2\sqrt{t}}=\sqrt{t}+c=\sqrt{x^2+\frac{1}{x^2}+1}+c=\frac{\sqrt{x^4+x^2+1}}{x}+c$