Question

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

The given integral is
$\int \frac{x^{2}-1}{x^{3} \sqrt{2 x^{4}-2 x^{2}+1}} d x$
$=\int \frac{x^{2}-1}{x^{5} \sqrt{2-\frac{2}{x^{2}}+\frac{1}{x^{4}}}} d x $
$=\int \frac{\left(\frac{1}{x^{3}}-\frac{1}{x^{5}}\right) d x}{\sqrt{2-\frac{2}{x^{2}}+\frac{1}{x^{4}}}}$
$=\frac{1}{2} \int \frac{t d t}{t},$
Let $2-\frac{2}{x^{2}}+\frac{1}{x^{4}}=t^{2}$
$\Rightarrow \left(\frac{2}{25}-\frac{4}{25}\right) d x=2 t d t$
$=(1 / 2) t+c$
$=\frac{1}{2} \sqrt{2-\frac{2}{x^{2}}+\frac{1}{x^{4}}}+c$