What is the antiderivative of lnx/x^2lnxx2?

1 Answer
Bdub
Apr 5, 2016

int(lnx)/x^2 =-(lnx+1)/x∫lnxx2=−lnx+1x

Explanation:

u=ln x, dv=1/x^2u=lnx,dv=1x2
du=1/x, v=-1/xdu=1x,v=−1x
intudv=uv-intvdu∫udv=uv−∫vdu
=-lnx/x-int-1/x^2 dx=−lnxx−∫−1x2dx
=-lnx/x+int1/x^2 dx=−lnxx+∫1x2dx
=-lnx/x-1/x=−lnxx−1x
=-(lnx+1)/x=−lnx+1x

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