Calculus/??

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Question
How to solve
Definite integral question...

Integrate [with limits from 0 to infinity] f(x)dx

where f(x) = l(x)/c(x)

and
l(x) = log(x)
and
c(x) = 1 + x^(1/3)

Please answer.
Thanks...

Answer
I  looked at that for awhile.    Eventually  I decided to integrate by part where
u = log(x) and dv = 1 + x^(1/3) dx.  That gives du = 1/x dx and v = x + (3/4)x^(4/3).

The answer is then uv - integral(v du).

That works out to be log(x)(x + (3/4)x^(4/3)) - integral((1/x)(x + (3/4)x^(4/3))).

Inside the integral, take (1/x)(x + (3/4)x^(4/3)) and multiply it out, giving
1 + (3/4)x^(1/3).  That integrates to x + (9/16)x^(4/3) + C.

This makes the answer be log(x)(x + (3/4)x^(4/3)) - x - (9/16)x^(4/3) + C.

Calculus

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