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Question

dxx5/6(1+5x1/3)1+4x1/3 equals

A
6tan1(1+4x1/3x1/3)+c
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B
3ln1+4+x1/314+x1/3+c
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C
6cot1(1+4x1/3x1/3)+c
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D
3ln14+x1/314x1/3+c
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Solution

The correct option is C 6cot1(1+4x1/3x1/3)+c
Put x1/3=t2
Given integration =6dt(1+5t2)1+4t2
Let 4+1t2=u2
I=6duu2+1=6cot14+1x1/3+c

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