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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Mathematics
Chapter
Indefinite Integration

JEE Advanced 2006 · Mathematics · Indefinite Integration

The value of ∫ (x^2-1) d x x^3 2 x^4-2 x^2+1 is

The value of ∫(x2−1)dxx32x4−2x2+1\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}} is

Options

  1. A22−2x2+1x4+C2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C
  2. B22+2x2+1x4+C2 \sqrt{2+\frac{2}{x^2}+\frac{1}{x^4}}+C
  3. C122−2x2+1x4+C\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C
  4. DNone of these

Correct answer

C. 122−2x2+1x4+C\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C

Step-by-step solution

Let I=∫(x2−1)dxx32x4−2x2+1\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 x5x^5, we get =∫(1x3−1x5)dx2−2x2+1x4\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 2−2x2+1x4=t⇒(4x3−4x5)dx=dt2-\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=14∫dtt=14⋅t1/21/2=12t+\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=122−2x2+1x4+\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C
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