Question

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

• A. $\frac{\sqrt{2x^4-2x^5+1}}{x^2}+C$
• 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 $\frac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$

Let $I=\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$

Dividing numerator and denominator by $x^5$, we get
$I=\int \frac{(\frac{1}{x^3}-\frac{1}{x^5})dx}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$

Substituting $2-\frac{2}{x^2}+\frac{1}{x^4}=t\Rightarrow (\frac{4}{x^3}-\frac{4}{x^5})dx=dt$

$I=\frac{1}{4}\int \frac{dt}{\sqrt{t}}=\frac{1}{4}.\frac{t^{1/2}}{1/2}+c$
$=\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$

$=\frac{\sqrt{2x^4-2x^2+1}}{2x^2}+c$