Answer
$= x(\ln x)^{2} - 2x\ln x + 2x +C $
Work Step by Step
$\int (\ln x)^{2}dx$
$u = (\ln x)^{2}$
$u' = 2(\ln x)(\frac{1}{x})$
$\frac{du}{dx} = 2(\ln x)(\frac{1}{x})$
$du = 2(\ln x)(\frac{1}{x})dx$
$dv = 1$
$v = x$
1. Find the integral
$uv - \int vdu$
$= ((\ln x)^{2})(x) - \int (x)2(\ln x)(\frac{1}{x})dx$
$= ((\ln x)^{2})(x) - \int 2x(\ln x)(\frac{1}{x})dx$
$= ((\ln x)^{2})(x) - \int (\ln x)(\frac{2x}{x})dx$
$= ((\ln x)^{2})(x) - \int (\ln x)2)dx$
$= ((\ln x)^{2})(x) - \int 2\ln (x) dx$
$= ((\ln x)^{2})(x) - 2\int \ln (x) dx$
$= ((\ln x)^{2})(x) - 2\int \ln (x) dx$
2. Find the integral of $\ln x$
$u = \ln x$
$u' = \frac{1}{x}$
$\frac{du}{dx} = \frac{1}{x}$
$du = \frac{dx}{x}$
$dv = 1$
$v = x$
$uv - \int vdu$
$= (\ln x)(x) - \int x\frac{1}{x}dx$
$= (\ln x)(x) - \int \frac{x}{x}dx$
$= (\ln x)(x) - \int 1dx$
$= (\ln x)(x) - x$
$= x(\ln x - 1)+C_1$
3. Substitute the integral you found in #2 back into the equation in #1 (Note: When we put the integral of $\ln x$ back into #1, we do not include $C$. Only at the end is when we put $C$)
$= ((\ln x)^{2})(x) - 2\int \ln (x) dx$
$= ((\ln x)^{2})(x) - 2[x(\ln x - 1)] +C $
$= ((\ln x)^{2})(x) - 2[x\ln x - x] +C $
$= ((\ln x)^{2})(x) - 2x\ln x + 2x +C $ (Where $C = 2C_1$)
$= x(\ln x)^{2} - 2x\ln x + 2x +C $ (Where $C = 2C_1$)
