## Calculus: Early Transcendentals 8th Edition

$= x(\ln x)^{2} - 2x\ln x + 2x +C$
$\int (\ln x)^{2}dx$ $u = (\ln x)^{2}$ $u' = 2(\ln x)(\frac{1}{x})$ $\frac{du}{dx} = 2(\ln x)(\frac{1}{x})$ $du = 2(\ln x)(\frac{1}{x})dx$ $dv = 1$ $v = x$ 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 $C = 2C_1$)