The value of \( I=\int(x^{3}-1)^{1/3}x^{5}\,dx \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{(x^{3}-1)^{7/3}}{7}+\displaystyle\frac{(x^{3}-1)^{4/3}}{4}+C \), where \( C \) is an arbitrary constant
Writing \( x^{5}=x^{3}\cdot x^{2} \), the substitution \( t=x^{3}-1 \) gives \( dt=3x^{2}\,dx \) and \( x^{3}=t+1 \):
\( \displaystyle I=\int(x^{3}-1)^{1/3}\,x^{3}\cdot x^{2}\,dx \)
\( \displaystyle =\frac{1}{3}\int t^{1/3}(t+1)\,dt \)
\( \displaystyle =\frac{1}{3}\int\left(t^{4/3}+t^{1/3}\right)dt \)
\( \displaystyle =\frac{1}{3}\left(\frac{3}{7}t^{7/3}+\frac{3}{4}t^{4/3}\right)+C \)
\( \displaystyle =\frac{t^{7/3}}{7}+\frac{t^{4/3}}{4}+C \)
\( \displaystyle =\frac{(x^{3}-1)^{7/3}}{7}+\frac{(x^{3}-1)^{4/3}}{4}+C \)