Question

$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}$ dx is equal to

• A. $\frac{\sqrt{2x^4-2x^2+1}}{x^2}+c$
• B. $\frac{\sqrt{2x^4-2x^2+1}}{x^3}+c$
• C. $\frac{\sqrt{2x^4-2x^2+1}}{x}+c$
• D. $\frac{\sqrt{2x^4-2x^2+1}}{2x^2}+c$

Answer 1

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

$I=\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$
$=\int \frac{(\frac{1}{x^3}-\frac{1}{x^5})dx}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$
Put
$2-\frac{2}{x^2}+\frac{1}{x^4}=z$
$\Rightarrow I=\frac{1}{4}\int \frac{dz}{\sqrt{z}}$
$=\frac{1}{2}\times \sqrt{z}+c\Rightarrow \frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$.