Question

The value of $\int \frac{(x^2-1)dx}{x^3\sqrt{2x^4-2x^2+1}}$ is

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

Answer 1

Answer: (C)

Let I$=\int \frac{(x^2-1)dx}{x^3\sqrt{2x^4-2x^2+1}}$
$[dividingnumeratorandenominatorby{x}^5]$
$=\int \frac{(\frac{1}{x^3}-\frac{1}{x^5})dx}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$
Put$2-\frac{2}{x^2}+\frac{1}{x^4}=t\Rightarrow (\frac{4}{x^3}-\frac{4}{x^5})dx=dt$
$\therefore I=\frac{1}{4}\int \frac{dt}{\sqrt{t}}=\frac{1}{4}.\frac{t^{1/2}}{1/2}+c$
$=\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$