What is $int (ln(x+1)/(x^2)) dx$ ?

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Answer 1 · 2018-04-12T22:48:25

$intln(x+1)/x^2dx=ln(x/(x+1))-ln(x+1)/x+"c"$

For $intln(x+1)/x^2dx$ , we integrate by parts

Let $u=ln(x+1)rArrdu=1/(x+1)dx$

and $dv=1/x^2dxrArrv=-1/x$

So

$intln(x+1)/x^2dx=-ln(x+1)/x+int1/(x(x+1))dx=int1/x-1/(x+1)dx-ln(x+1)/x=lnx-ln(x+1)-ln(x+1)/x$

What is int (ln(x+1)/(x^2)) dxint (ln(x+1)/(x^2)) dx?