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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
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