## Calculus: Early Transcendentals 8th Edition

$$\displaystyle\int_{1}^{2} \frac{(\ln x)^{2}}{x^{3}}\thinspace dx =\frac{3}{16}-\frac{1}{8}\big((\ln 2)^{2}+\ln 2)$$
$$\displaystyle\int_{1}^{2} \frac{(\ln x)^{2}}{x^{3}}\thinspace dx$$ To solve this integral, you need to use integration by parts; once you do that, you will need to do integration by parts one more time. $u = (\ln x)^{2}$ $du = 2\ln (x\times \frac{1}{x})$ $dv = \frac{1}{x^{3}}$ $v = \frac{-1}{2x^{2}}$ $\displaystyle\int_{1}^{2} \frac{(\ln x)^{2}}{x^{3}}\thinspace dx = (\ln x)^{2}\times \big(\frac{-1}{2x^{2}}\big) - \displaystyle\int_{1}^{2} \frac{-1}{2x^{2}} \times \frac{2\ln x}{x}\thinspace dx - \frac{(\ln x)^{2}}{2x^{2}}+\displaystyle\int_{1}^{2} \frac{\ln x}{x^{3}}\thinspace dx$ The second time: $u = \ln x$ $\quad$ $du = \frac{1}{2}$ $dv = \frac{1}{x^{3}} \thinspace dx$ $\quad$ $v = \frac{-1}{2x^{2}}$ $\displaystyle\int_{1}^{2} \frac{(\ln x)^{2}}{x^{3}}\thinspace dx = -\frac{(\ln x)^{2}}{2x^{2}} + \ln x \times \frac{-1}{2x^{2}} - \displaystyle\int_{1}^{2} \frac{-1}{2x^{2}} \times \frac{1}{x} \thinspace dx$ $$= -\frac{(\ln x)^{2}}{2x^{2}} + \ln x \times\frac{-1}{2x^{2}}+\frac{1}{2}\displaystyle\int_{1}^{2} \frac{1}{x^{3}}\thinspace dx$$ $$-\frac{(\ln x)^{2}}{2x^{2}}+\ln x \times \frac{-1}{2x^{2}}+\frac{1}{2}\times\frac{-1}{2x^{2}}\bigg|^{2}_1$$ $$=-\frac{(\ln x)^{2}}{2x^{2}}-\frac{\ln x}{2x^{2}}-\frac{1}{4x^{2}}\bigg|^{2}_1$$ $$=-\Bigg(\frac{2(\ln x)^{2}+2\ln x +1}{4x^{2}}\Bigg)\Bigg|^{2}_1$$ $$=-\Bigg(\frac{2(\ln 2)^{2}+2\ln 2+1}{4\times2^{2}}\Bigg)+\Bigg(\frac{2(\ln 1)^{2}+2\ln 1+1}{4\times1^{2}}\Bigg)$$ $$=-\Bigg(\frac{2(\ln 2)^{2} +2\ln 2+1}{16}\Bigg)+\frac{1}{4}$$ $$=\frac{3}{16}-\frac{1}{8}\big((\ln2)^{2}+\ln 2)\big)$$