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AIEEE 2006

JEE Main 2006 Mathematics question on Indefinite Integration

The value of ∫(x2−1) dxx32x4−2x2+1\int \dfrac{(x^2-1)\,dx}{x^3\sqrt{2x^4-2x^2+1}} is \hfill
  1. A.
    22−2x2+1x4+c2\sqrt{2-\dfrac{2}{x^2}+\dfrac{1}{x^4}}+c
  2. B.
    2+2x2+1x4+c\sqrt{2+\dfrac{2}{x^2}+\dfrac{1}{x^4}}+c
  3. C.
    122−2x2+1x4+c\dfrac{1}{2}\sqrt{2-\dfrac{2}{x^2}+\dfrac{1}{x^4}}+c
  4. D.
    None of these
Show answer and solution

Correct option: C

Solution

Divide numerator and denominator by x⁴. The integral becomes:

∫ (x²-1)/x³ · dx / √(2x⁴-2x²+1)

Divide num and denom by x⁴:

Numerator: (x²-1)/x³ · (1/x) = (x²-1)/x⁴

Denominator: √(2x⁴-2x²+1)/x⁴ = √((2x⁴-2x²+1)/x⁸) = √(2/x⁴ - 2/x⁶ + 1/x⁸)...

Better approach: rewrite 2x⁴-2x²+1 = x⁴(2 - 2/x² + 1/x⁴), so √(2x⁴-2x²+1) = x²√(2 - 2/x² + 1/x⁴).

The integral becomes: ∫ (x²-1)/(x³ · x²√(2-2/x²+1/x⁴)) dx = ∫ (x²-1)/(x⁵√(2-2/x²+1/x⁴)) dx

Now let t = 2 - 2/x² + 1/x⁴. Then dt/dx = 4/x³ - 4/x⁵ = 4(x²-1)/x⁵.

So (x²-1)/x⁵ dx = dt/4.

The integral becomes: ∫

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