Question

The value of \[ \int_{0}^{\infty}\frac{x^{2}(1-x^{2})}{(1+x)^{24}}\,dx \] is

• A. \(0\)
• B. \(8\)
• C. \(24\)
• D. \(1\)

Hint:

For integrals over \((0,\infty)\), \[ \boxed{x\rightarrow\frac1x} \] is a useful substitution. If the integrand changes sign under this transformation, then the integral is \[ \boxed{0.} \]

Answer 1

The Correct Option is A

Step 1: Let \[ I=\int_{0}^{\infty}\frac{x^{2}(1-x^{2})}{(1+x)^{24}}\,dx. \] Apply the substitution \[ x=\frac1t,\qquad dx=-\frac{dt}{t^{2}}. \] Then \[ I=\int_{0}^{\infty} \frac{t^{20}(t^{2}-1)}{(1+t)^{24}}\,dt. \]

Step 2: Add the two expressions for \(I\). Adding, \[ 2I= \int_{0}^{\infty} \frac{(1-t^{2})(t^{2}+t^{20})}{(1+t)^{24}}\,dt. \] Since the transformed integrand is the negative of the original under the substitution \(x\mapsto1/x\), \[ I=-I. \] Hence, \[ 2I=0 \] which gives \[ \boxed{I=0.} \] Therefore, \[ \boxed{(A)} \] is the correct answer.