Find # I = int \ lnx/x^2 \ dx #?
1 Answer
Oct 11, 2017
# int \ (lnx)/x^2 \ dx = -(lnx+1)/x + C #
Explanation:
We seek:
# I = int \ lnx/x^2 \ dx #
We can apply integration by 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 #
gives us
# int \ (lnx)(1/x^2) \ dx = (lnx)(-1/x) - int \ (-1/x)(1/x) \ dx #
# :. int \ (lnx)/x^2 \ dx = -(lnx)/x + int \ 1/x^2 \ dx #
# :. int \ (lnx)/x^2 \ dx = -(lnx)/x - 1/x + C #
# :. int \ (lnx)/x^2 \ dx = -(lnx+1)/x + C #
