Help me evaluate this integral

Question

$\displaystyle \int \frac{\ln x}{x^2} \mathrm dx$

I just can't seem to figure this one out. I tried integrating by parts but I'm stuck.

Answer 1

Integration by parts is not u-substitution. You split the integrand into two parts, $u$ and $dv$. Choose $u$ so that it will simplify when you take it's derivative. In this case, $u$ simplifies when it is ln(x) and the derivative is 1/x .

$u = \ln(x)$

$du = (1/x)dx$

$v = -1/x$

$dv = (1/(x^2))dx$

$uv - \int{vdu} = (\ln(x))\cdot(-1/x)- \int{(-1/(x^2)) dx} = (-\ln x)/x - (1/x) = -( ln x+1)/x$