Show that integral of

∫ dx / [ (x+1) ^1/3 + (x+1) ^1/2 ] = 2t³ - 3t² + 6t - log|1+t| +c where t = (x+1) ^1/6

if t=(x+1)^1/6   =>  dt=1/6(x+1)^-5/6 dx

 =>dx=6(x+1)^5/6dt

now the ques

 6 ∫ [t⁵dt] / [t²+t³]  =>    6 ∫ [t³dt] / [1+t]

 now dividing the integral into two we get

 6[ [∫{t³+1}dt / (t+1) ]- [∫dt/(t+1)]  ]

6 [ [∫(t²+1-t)dt]-log|1+t|  ]    =>  I= 2t³ - 3t²+ 6t- 6log|1+t| + c