\[\int\frac{1}{ln(x+1)} dx\]

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\[\int\frac{1}{ln(x+1)} dx\]

Mathematics
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lemme try
Hint: similar to your other one.

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Other answers:

But this is apparently an integral which does not have a close form..
okay
ash dont solve its for me
I ain't solving, just for reference http://www.wolframalpha.com/input/?i=integral+1%2F%28ln+%28x%2B1%29%29+dx
this is like wasiqs previous integral which was apparently solveable http://www.wolframalpha.com/input/?i=integral+e^x%2F%28x%29
mishoo it was easy ~~~ i did it
just apply by parts
by parts? please elaborate.
i thought the first step is a substitution..
yeh take 1 as first functio
\[x/{(\ln(x+1)} - \int\limits_{}^{}x^2 +x\]
how about \(u= ln(x+1)\) => \(e^{u}= x+1\)
if it can be done easily why take substitution..
but this is an integral which does not have a closed form right?
i dont understand forms :P i just know how to solve the ugliest of integrals :P
lol, but look at the answer on wolfram..
i dont trust wolfram cause in two questions i pointed out its mistake and did what it cant do :P
lol are you sure that you're right though? :P
i can do better than wolfram :P
lol

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