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How to solve this integration

QQ20170726-233531@2x.png
Original post by ElliotWalton
QQ20170726-233531@2x.png


Where did you stumble across this? Are you aware that there's no closed form?
Original post by _gcx
Where did you stumble across this? Are you aware that there's no closed form?


this is an inside interesting integrals
Original post by ElliotWalton
this is an inside interesting integrals


If you do want help with it. You'll need to know that,

−Li2(−x)=∫ln⁡(x+1)xdx\displaystyle -\mathrm{Li}_2(-x) = \int \frac{\ln(x+1)}{x} \mathrm{d}x

How can you rewrite ln⁡(x+1)(1x2+x)\displaystyle \ln(x+1)\left(\frac{1}{x^2 + x}\right)?
Original post by RDKGames
∫ln⁡(1+x)x2+x.dx=∫x−1⋅ln⁡(1+x)x.dx\displaystyle \int \frac{\ln(1+x)}{x^2+x} .dx = \int x^{-1}\cdot \frac{\ln(1+x)}{x} .dx then IBP


can you write down all the workings?
Original post by ElliotWalton
can you write down all the workings?


Are you sure that what you write is correct?
Original post by AnIndianGuy
He's right, trust me


I have no ideas distinguishing whether He is correct or not.I hope you guys can show a specific working.THX
Original post by RDKGames
∫ln⁡(1+x)x2+x.dx=∫x−1⋅ln⁡(1+x)x.dx\displaystyle \int \frac{\ln(1+x)}{x^2+x} .dx = \int x^{-1}\cdot \frac{\ln(1+x)}{x} .dx then IBP


I was trying to get onto something different, ie.

Spoiler

Original post by _gcx
I was trying to get onto something different, ie.

Spoiler




I misread the Q when it came to the workings

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