What is #int (1+lnx)/x^2dx#?

1 Answer
Mar 15, 2018

Answer:

# 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 #