Integrals: Edge Cases and Subtle Traps (7)

Question

Evaluate $\displaystyle\int_{-1}^{1} \frac{x^3}{1+x^2}\,dx$.

Solution — Step by Step

The integral equals $0$.

Why This Works

An odd function has equal positive and negative areas on a symmetric interval, so they cancel. We don’t need to find an antiderivative — the symmetry argument is a one-line solution.

This is one of the most reliable shortcuts in JEE. Whenever you see an integral on $[-a, a]$, immediately check parity. If odd, the answer is zero. If even, the answer is twice the integral on $[0, a]$.

Alternative Method

Substitute $u = 1 + x^2$, so $du = 2x\,dx$ and $x^2 = u - 1$.

$\int \frac{x^3}{1+x^2}\,dx = \int \frac{x^2 \cdot x}{1+x^2}\,dx = \frac{1}{2}\int \frac{u-1}{u}\,du = \frac{1}{2}(u - \ln|u|) + C$

Evaluating between $-1$ and $1$ gives $0$, since the antiderivative is even (depends on $x^2$). Same answer, more work.

Common Mistake

Skipping the parity check and grinding through the substitution. You waste five minutes and risk an arithmetic error. The fix: every time the integral is on $[-a, a]$, make parity check your first step.