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

1 Answer

Apr 12, 2018

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

Explanation:

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#

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