How do you integrate #int lnx/x^2# by integration by parts method?
1 Answer
Apr 28, 2018
# int \ (lnx)/x^2 \ dx = -(1+lnx)/x+C#
Explanation:
We seek:
# I = int \ (lnx)/x^2 \ dx #
We can then apply Integration By Parts:
Let
# { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x^2, => v,=-1/x ) :}#
Then plugging into the IBP formula:
# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #
We have:
# int \ (lnx)(1/x^2) \ dx = (lnx)(-1/x) - int \ (-1/x)(1/x) \ dx #
# :. I = -(lnx)/x + int \ 1/x^2 \ dx #
# \ \ \ \ \ \ \ = -(lnx)/x -1/x + C #
# \ \ \ \ \ \ \ = -(1+lnx)/x + C #
