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If dx1+x+13=32(x+1)2/33(x+1)1/3

+f(x)+C then f(x) is equal to

 

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a
log⁡|1+x+13|
b
3log⁡|1+x+13|
c
23log⁡|1+x+13|
d
13log⁡|1+x+13|

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detailed solution

Correct option is B

Put x+1=t3, we have∫dx1+x+13=∫3t2dt1+t=3∫t−1+11+tdt=3t22−t+log⁡|1+t|+C=32(x+1)2/3−3(x+1)1/3+3log⁡|1+x+13|+C


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