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\int \frac{ln(x+1)}{x^2}dx

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u=x+1

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\int\frac{lnu}{(u-1)^2}du

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\int \frac{A}{u-1}+\frac{B}{(u-1)^2}du

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lnu=A(u-1)+B

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B=0,A=ln2

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\int \frac{ln2}{u-1}du

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ln2ln|x|+C

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