Q43 / 120 · JEE Advanced 2006 · Mathematics
The value of $\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}$ is
- A$2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$
- B$2 \sqrt{2+\frac{2}{x^2}+\frac{1}{x^4}}+C$
- Correct answerC$\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$
- DNone of these
Solution
Let $\quad I=\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}$,
On dividing numerater and denominator by $x^5$, we get
$$
=\int \frac{\left(\frac{1}{x^3}-\frac{1}{x^5}\right) d x}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}
$$
put $2-\frac{2}{x^2}+\frac{1}{x^4}=t \Rightarrow\left(\frac{4}{x^3}-\frac{4}{x^5}\right) d x=d t$
$$
\therefore \quad I=\frac{1}{4} \int \frac{d t}{\sqrt{t}}=\frac{1}{4} \cdot \frac{t^{1 / 2}}{1 / 2}=\frac{1}{2} \sqrt{t}+c=\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C
$$
Hence, the correct answer is option: 'C'.