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Show that integral of ∫ dx / [ (x+1) 1/3 + (x+1) 1/2 ] = 2t 3 - 3t 2 + 6t - log|1+t| +c where t = (x+1) 1/6

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

Grade:12

1 Answers

Vanya Saxena
18 Points
11 years ago
 

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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