How do you integrate #int lnx/x^2# by integration by parts method?

Question

Charles Anctil · Accepted Answer

# int \ (lnx)/x^2 \ dx = -(1+lnx)/x+C#

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 #

Isabella Ackerman · Answer

undefined To integrate $ \int \frac{\ln(x)}{x^2} $ using integration by parts, you can choose $ u = \ln(x) $ and $ dv = \frac{1}{x^2} \, dx $.

This yields $ du = \frac{1}{x} \, dx $ and $ v = -\frac{1}{x} $.

Applying the integration by parts formula, $ \int u \, dv = uv - \int v \, du $, we get:

$ \int \frac{\ln(x)}{x^2} \, dx = -\frac{\ln(x)}{x} + \int \frac{1}{x^2} \, dx $

The second term on the right-hand side can be integrated directly. So, the final result is:

$ \int \frac{\ln(x)}{x^2} \, dx = -\frac{\ln(x)}{x} - \frac{1}{x} + C $

Where $ C $ is the constant of integration.