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JEE Advanced · Mathematics · 26. Indefinite Integration

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

  1. A
    \(2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C\)
  2. B
    \(2 \sqrt{2+\frac{2}{x^2}+\frac{1}{x^4}}+C\)
  3. C
    \(\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C\)
  4. D
    None of these
Verified Solution

Answer & Solution

Correct Answer

(C)
\(\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C\)

Step-by-step Solution

Detailed explanation

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 \(2-\frac{2}{x^2}+\frac{1}{x^4}=t \Rightarrow\left(\frac{4}{x^3}-\frac{4}{x^5}\right) d x=d t\)
\[
\therefore \quad I=\frac{1}{4} \int \frac{d t}{\sqrt{t}}=\frac{1}{4} \cdot \frac{t^{1 / 2}}{1 / 2}=\frac{1}{2} \sqrt{t}+c=\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C
\]
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