∫ (x^8)/(≤ft(x^3+1)^1/3) dx is equal to

The correct answer is Option (4) → \( \displaystyle\frac{1}{8}\left(x^{3}+1\right)^{8/3}+\frac{1}{2}\left(x^{3}+1\right)^{2/3}-\frac{2}{5}\left(x^{3}+1\right)^{5/3}+C \), C is an arbitrary constant

Let \( t=x^{3}+1 \) (a substitution). Then \( 3x^{2}\,dx=dt \) and \( x^{3}=t-1 \).

Split \( x^{8}=(x^{3})^{2}\cdot x^{2} \), so that \( x^{2}\,dx \) becomes \( \frac{dt}{3} \):

    \( \displaystyle \int\frac{x^{8}}{(x^{3}+1)^{1/3}}\,dx=\int\frac{(x^{3})^{2}\cdot x^{2}\,dx}{(x^{3}+1)^{1/3}} \)

    \( \displaystyle =\frac{1}{3}\int\frac{(t-1)^{2}}{t^{1/3}}\,dt \)

Expanding \( (t-1)^{2}=t^{2}-2t+1 \) and dividing each term by \( t^{1/3} \):

    \( \displaystyle =\frac{1}{3}\int\left(t^{5/3}-2t^{2/3}+t^{-1/3}\right)dt \)

    \( \displaystyle =\frac{1}{3}\left[\frac{3}{8}t^{8/3}-2\cdot\frac{3}{5}t^{5/3}+\frac{3}{2}t^{2/3}\right]+C \)

    \( \displaystyle =\frac{1}{8}t^{8/3}-\frac{2}{5}t^{5/3}+\frac{1}{2}t^{2/3}+C \)

Putting back \( t=x^{3}+1 \):

    \( \displaystyle =\frac{1}{8}\left(x^{3}+1\right)^{8/3}+\frac{1}{2}\left(x^{3}+1\right)^{2/3}-\frac{2}{5}\left(x^{3}+1\right)^{5/3}+C \)

This is Option (4), with the terms in a different order.