\(\int\displaystyle\frac{(x^4 - x)^{1/4}}{x^5}\, dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{5/4} + C\): \(C\) is a constant of integration
Taking \(x^4\) out of the bracket:
\( \displaystyle (x^4 - x)^{1/4} = \left[x^4\left(1 - \frac{1}{x^3}\right)\right]^{1/4} = x\left(1 - \frac{1}{x^3}\right)^{1/4} \)
So the integral becomes
\( \displaystyle I = \int \frac{1}{x^4}\left(1 - \frac{1}{x^3}\right)^{1/4} dx \)
By substitution, let \(t = 1 - x^{-3}\):
\( \displaystyle dt = 3x^{-4}\,dx,\quad \frac{dx}{x^4} = \frac{dt}{3} \)
\( \displaystyle I = \frac{1}{3}\int t^{1/4}\,dt = \frac{1}{3} \cdot \frac{t^{5/4}}{5/4} + C \)
\( \displaystyle = \frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{5/4} + C \)