Vtyjkyuk

Study the convergence of the following integral for $alphainmathbbR$ and evaluate it for $alpha = frac13$ if it converges for that value.

$$Largeint _0^1fracdx(e^x-1)^alpha x$$

The integral converges for $alpha lt 1$, if I am not wrong, so I can evaluate:
$$Largeint _0^1fracdx(e^x-1)^frac13$$

whose primitive, using some razionalizations and change of variables, is: $frac12log|4(e^x-1)^frac23-4(e^x-1)^frac13+4|+sqrt3arctanleft(frac1sqrt3left(2(e^x-1)^frac13-1right)right)-log|(e^x-1)^frac13+1|+C$

Anyone knows if there is a less intricate way in order of obtain it than almost 3 sheet full of calculations? Thank you

$$int_0^1fracdxsqrt[3]e^x-1stackrelxmapsto log t=int_1^efracdttsqrt[3]t-1stackreltmapsto s+1=int_0^e-1fracds(s-1)sqrt[3]sstackrelsmapsto u^3=int_0^sqrt[3]e-1frac3uu^3-1,du$$
where
$$ frac3uu^3-1 stackreltextResidues= frac1u-1+frac1-uu^2+u+1 $$
and
$$ frac1-uu^2+u+1 = frac4(2u+1)^2+3-frac4u(2u+1)^2+3 $$
lead to the answer in less than a page. This is a trick you are probably going to apply many times: turn the integrand function into a rational function, then perform a partial fraction decomposition. Objects like $frac1(u+alpha)^n$ and $frac1(u^2+alpha^2)^n$ always have simple primitives, that's the core of symbolic integration algorithms.

using $u=left( e^x-1 right)^1/3$ leads to $intfrac3uu^3+1du$
$$
beginalign
& =int-frac1u+1+fracu+1u^2-u+1du \
& =int-frac1u+1du+frac12intfrac2u-1u^2-u+1du+frac32intfrac1u^2-u+1du \
& =ln left( u+1 right)+frac12ln left( u^2-u+1 right)+frac32intfrac1u^2-u+1du \
endalign
$$
For the last integral use $v=2u-1$ and notice that $4left( u^2-u+1 right)=v^2+3$
$$
beginalign
& =ln left( u+1 right)+frac12ln left( u^2-u+1 right)+3intfrac1v^2+3dv \
& =ln left( u+1 right)+frac12ln left( u^2-u+1 right)+3frac1sqrt3tan^-1left( frac2u-1sqrt3 right) \
endalign
$$

Take
$$t = (e^x - 1)^1/3$$
or
$$x = ln( t^3 + 1 )$$
Then
$$dt = frac13e^x(e^x-1)^-2/3 dx$$
Hence
$$Largeint _0^1fracdx(e^x-1)^frac13
=
3Largeint _0^(e-1)^1/3frac1 t e^x(e^x-1)^-2/3 dt
=
3Largeint _0^(e-1)^1/3fract(t^3 + 1) dt
$$
The last integral could be written as
$$
displaystyleint _0^(e-1)^1/3dfractleft(t+1right)left(t^2-t+1right),mathrmdt$$
Perform Partial Fraction Decomposition
$$displaystyleint _0^(e-1)^1/3left(dfract+13left(t^2-t+1right)-dfrac13left(t+1right)right)mathrmdt$$
Take the first integral above, i.e. $displaystyleintdfract+1t^2-t+1,mathrmdt$ could be written as
$$classsteps-nodecssIdsteps-node-7dfrac12displaystyleintdfrac2t-1t^2-t+1,mathrmdt+classsteps-nodecssIdsteps-node-8dfrac32displaystyleintdfrac1t^2-t+1,mathrmdt$$
which is equal to (with some straight forward math)
$$dfraclnleft(t^2-t+1right)2+sqrt3arctanleft(dfrac2t-1sqrt3right)$$
Now, we need $dfraclnleft(t^2-t+1right)2+sqrt3arctanleft(dfrac2t-1sqrt3right)$
which is easily shown to be
$$dfraclnleft(t^2-t+1right)2+sqrt3arctanleft(dfrac2t-1sqrt3right)==lnleft(t+1right)$$
Finalizing we get
$$3left(-dfracright)3+dfraclnleft(t^2-t+1right)6+dfracarctanleft(frac2t-1sqrt3right)sqrt3right)Bigvert_0^(e-1)^1/3$$
which is
$$3left(dfraclnleft(left(mathrme-1right)^frac23-sqrt[3]mathrme-1+1right)6+dfracarctanleft(frac2cdotsqrt3sqrt[3]mathrme-1-sqrt33right)sqrt3-dfraclnleft(sqrt[3]mathrme-1+1right)3+dfracpi2cdot3^frac32right)$$