What is #int (1+lnx)/x^2dx#?
1 Answer
Mar 15, 2018
# int \ (1+lnx)/x^2 \ dx = -(2+lnx)/x C #
Explanation:
We seek:
# I = int \ (1+lnx)/x^2 \ dx #
In preparation for an application of [integration by Parts]We can then apply Integration By Parts:
Let
# { (u,=1+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 \ (1+lnx)(1/x^2) \ dx = (1+lnx)(-1/x) - int \ (-1/x)(1/x) \ dx #
Giving is:
# I = -(1+lnx)/x + int \ 1/x^2 \ dx #
# \ \ = -(1+lnx)/x -1/x + C #
# \ \ = -(2+lnx)/x C #