$∫\frac{dx}{\sqrt{x+1}+(x+1)^{\frac{1}{3}}}$ is equal to

$2p^3-3P^2+6p-6ln(1+p)+c$, where $p=(x+1)^{\frac{1}{6}}$ $2p^3+6p-6ln(1+p)+c$, where $p=(x+1)^{\frac{1}{6}}$ $2p^3+3P^2+6p-6ln(1+p)+c$, where $p=(x+1)^{\frac{1}{6}}$ none of these

$2p^3-3P^2+6p-6ln(1+p)+c$, where $p=(x+1)^{\frac{1}{6}}$

Here $I=∫\frac{6p^5dp}{p^3+p^2}$  ((x + 1) = p6 , then dx = 6p5 dp). $=6∫p^2dp-6∫p\,dp+6∫dp-6∫\frac{p^2}{p^2(p+1)}dp$ $=\frac{6p^3}{3}-\frac{6p^2}{2}+6p-6ln(p+1)+c$ $=2p^3-3P^2+6p-6ln(1+p)+c$, where $p=(x+1)^{\frac{1}{6}}$ Hence (A) is the correct answer.