Q.Choose the correct answer: The value of the integral $\int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx$ is (A) 6 (B) 0 (C) 3 (D) 4

Concept: Definite Integral Symmetry — the integrand may simplify under a substitution that reveals a self-similar or symmetric form.

Step 1: Write the integral as

Step 2: Substitute $x = 1/t$, so $dx = -dt/t^2$. When $x = 1/3$, $t = 3$; when $x = 1$, $t = 1$. Then

Step 3: Simplify: $(1 - 1/t^2)^{1/3} = (t^2 - 1)^{1/3} / t^{2/3}$, and $1/t^{-11/3} = t^{11/3}$. So

$I = \int_{1}^{3} \frac{(t^2 - 1)^{1/3}}{t^{2/3}} \cdot t^{11/3} \cdot \frac{dt}{t^2} = \int_{1}^{3} (t^2 - 1)^{1/3} \cdot t^{11/3 - 2/3 - 2}\,dt = \int_{1}^{3} (t^2 - 1)^{1/3} \cdot t^{1}\,dt.$ …

We evaluate $\displaystyle\int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx$.

1. Factor inside the cube root. Since $x-x^3=x^3\!\left(\dfrac{1}{x^2}-1\right)$,

Therefore the integrand becomes

$\frac{x\left(\tfrac{1}{x^2}-1\right)^{1/3}}{x^4}=\frac{\left(\tfrac{1}{x^2}-1\right)^{1/3}}{x^3}.$ …

Method: Reciprocal substitution to expose a hidden power/derivative structure

For a messy algebraic integrand with high powers of $x$ in the denominator, the substitution $x=\tfrac{1}{t}$ (or factoring $x$ out of the root) often reveals a simple $t\,(t^2-1)^{k}$ shape that a second substitution finishes.

Steps

Step 1: Factor inside the root to isolate a power of $x$.

Write, e.g., $x - x^3 = x^3\big(\tfrac{1}{x^2}-1\big)$ so the root becomes $x\big(\tfrac{1}{x^2}-1\big)^{1/3}$; the awkward $x$-powers then cancel against the denominator.

Step 2: Substitute the leftover inner expression. …

Common Mistakes

Mistake 1: Trying to expand $(x-x^3)^{1/3}$ directly.

Why it's wrong: a cube root of a difference does not distribute; you must factor out $x^3$ (or $x$) first to expose a clean inner function. Correct approach: write $(x-x^3)^{1/3}=x\left(\tfrac{1}{x^2}-1\right)^{1/3}$.

Mistake 2: Mismatching the substitution's differential with the leftover $x$-power. …