Question:medium

If \(\displaystyle \int \frac{x\,dx}{\sqrt[15]{(1+x^2)^{12}(2+x^2)^{18}}}=\alpha\left(\frac{1+x^2}{2+x^2}\right)^{1/n}+C\), then \(\dfrac{n}{\alpha}=\)

Show Hint

When the integrand contains powers of \(1+x^2\) and \(2+x^2\), try the substitution \[ u=\frac{1+x^2}{2+x^2}. \] This substitution is useful because its derivative contains \[ \frac{x\,dx}{(2+x^2)^2}. \]
Updated On: Jun 18, 2026
  • \(6\)
  • \(4\)
  • \(2\)
  • \(8\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Rewrite the radical in exponent form.
The 15th root becomes exponent 1/15: (1+x²)^(12/15) (2+x²)^(18/15) = (1+x²)^(4/5) (2+x²)^(6/5). Integral = ∫ x dx / [(1+x²)^(4/5) (2+x²)^(6/5)].

Step 2: Factor and substitute cleverly.

Write as ∫ x dx / [(2+x²)² ((1+x²)/(2+x²))^(4/5)]. Let t = (1+x²)/(2+x²). Then dt = [2x(2+x²) – 2x(1+x²)]/(2+x²)² dx = 2x/(2+x²)² dx. So x dx/(2+x²)² = dt/2.

Step 3: Simplify and integrate.

Integral becomes ∫ (1/t^(4/5)) · (dt/2) = (1/2) ∫ t^(–4/5) dt = (1/2)·5·t^(1/5) + C = (5/2) t^(1/5) + C.

Step 4: Back-substitute and compare.

Result = (5/2)[(1+x²)/(2+x²)]^(1/5) + C. Comparing with α[(1+x²)/(2+x²)]^(1/n) gives α = 5/2, n = 5.

Step 5: Final Answer:

n/α = 5/(5/2) = 2.
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