Please help me solve the integral I = ln(x)/(x^2) dx Thank you
$\displaystyle \int ln(x)x^{-2}dx$
Integration by parts.
$\displaystyle \int f(x)g'(x)dx = f(x)g(x)-\int f'(x)g(x)dx$
Using the substitutions:
$\displaystyle f(x) = ln(x)$
$\displaystyle f'(x) = \frac{1}{x}$
$\displaystyle g(x) = -\frac{1}{x}$
$\displaystyle g'(x) = x^{-2}$
Which ends up as:
$\displaystyle -\frac{ln(x)}{x}+\int \frac{1}{x^2}dx$
$\displaystyle -\frac{ln(x)}{x}-\frac{1}{x} + C$
$\displaystyle -\frac{ln(x)}{x}-\frac{1}{x} \right )+C$
Post #3 is not correct. See the solution here.