\( \int_{1/3}^{1}\displaystyle\frac{(x-x^3)^{1/3}}{x^4}\,dx \) equals
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6
Since \( x-x^3=x^3\left(\frac{1}{x^2}-1\right) \), the bracket can be written with \( x^3 \) outside:
\( \displaystyle (x-x^3)^{1/3}=x\left(\frac{1}{x^2}-1\right)^{1/3} \)
\( \displaystyle \frac{(x-x^3)^{1/3}}{x^4}=\frac{1}{x^3}\left(\frac{1}{x^2}-1\right)^{1/3} \)
Let \( t=\frac{1}{x^2}-1 \) (substitution). Then
\( \displaystyle dt=-\frac{2}{x^3}\,dx \)
\( \displaystyle \frac{dx}{x^3}=-\frac{dt}{2} \)
The limits change: x = 1/3 gives t = 9 − 1 = 8, and x = 1 gives t = 0.
\( \displaystyle \int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx=-\frac{1}{2}\int_{8}^{0}t^{1/3}\,dt \)
\( \displaystyle =\frac{1}{2}\int_{0}^{8}t^{1/3}\,dt \)
\( \displaystyle =\frac{1}{2}\cdot\frac{3}{4}\Big[t^{4/3}\Big]_{0}^{8} \)
\( \displaystyle =\frac{3}{8}\times 16=6 \)