The value of \(\int\displaystyle\frac{(x^4-x)^{1/4}}{x^5}\,dx\) is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{5/4}+C\)
Taking \(x^4\) out of the bracket:
\( \displaystyle (x^4 - x)^{1/4} = x\left(1 - \frac{1}{x^3}\right)^{1/4} \)
\( \displaystyle \frac{(x^4-x)^{1/4}}{x^5} = \frac{1}{x^4}\left(1 - \frac{1}{x^3}\right)^{1/4} \)
The factor \(\frac{1}{x^4}\) is (up to a constant) the derivative of the bracket, so substitute
\( \displaystyle t = 1 - \frac{1}{x^3} \)
\( \displaystyle dt = \frac{3}{x^4}\,dx \)
Then
\( \displaystyle \int \frac{(x^4-x)^{1/4}}{x^5}\,dx = \frac{1}{3}\int t^{1/4}\,dt \)
\( \displaystyle = \frac{1}{3}\cdot\frac{4}{5}\,t^{5/4} + C \)
\( \displaystyle = \frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{5/4} + C \)
NCERT (2026-27 edition):
Integrals
