How do you find the integral # ln(x)/(x^2)#?

Question

How do you find the integral # ln(x)/(x^2)#?

1 Answer

Impact of this question

2428 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-07T11:12:01

#-lnx /x -1/x +C#

Explanation:

Integration by parts can be done in this case, #int (ln x)/x^2 dx# =#int lnx*d/dx(-1/x) dx#= #-lnx /x -int d/dx (lnx)*(-1)/x dx+C#= #-lnx /x+int 1/x^2 dx +C#= #-lnx /x -1/x +C#

Science

Math

Humanities

... and beyond

How do you find the integral # ln(x)/(x^2)#?

1 Answer

Explanation:

Related questions

Impact of this question