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
