Square Logs

$\begin{aligned} G(x) & = \int (x \ln x)^2 \, dx \quad \quad \small \color{#3D99F6}{\text{By integration by parts: } u = \ln^2 x, \, dv = x^2 dx} \\ & = \frac{x^3 \ln^2 x}{3} - \int \frac{x^3\cdot 2\ln x}{3x} \, dx \quad \quad \small \color{#3D99F6}{\text{By integration by parts again: } u = \ln x, \, dv = x^2 dx} \\ & = \frac{x^3 \ln^2 x}{3} - \frac{2 x^3 \ln x}{9} + \frac{2}{9} \int \frac{x^3}{x} \, dx \\ & = \frac{x^3 \ln^2 x}{3} - \frac{2 x^3 \ln x}{9} + \frac{2x^3}{27} + \color{#3D99F6}{c} \quad \quad \small \color{#3D99F6}{c = \text{constant}} \\ & = \frac{x^3}{27} (9 \ln^2 x - 6 \ln x + 2) + c \\ \implies G(e) & = \frac{e^3}{27} (9 - 6 + 2) + c \\ \frac{5e^3}{27} - 1 & = \frac{5e^3}{27} + c \\ \implies c & = -1 \\ \implies G(x) & = \frac{x^3}{27} (9 \ln^2 x - 6 \ln x + 2) - 1 \\ G(13) & \approx \boxed{3727.47} \end{aligned}$

You can solve by using integration by parts twice

$\int u \, dw =uw - \int w \, du$

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$G(x) = \int x^{2} \cdot ln(x)^{2}\,dx \;$ make the substitutions

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$u = ln(x)^{2}, \; du = 2x^{-1} \cdot ln(x) \, dx$

$dw = x^{2} \, dx, \; w = x^{3}/3$

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$G(x) = \int x^{2} \cdot ln(x)^{2}\,dx = \frac{x^{3}}{3}ln(x)^{2} - \frac{2}{3} \int x^{2} \cdot ln(x) \, dx$

Now to use integration by parts the second time. u(x) will be redefined

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$u = ln(x), \; du = x^{-1} \, dx$

$dw = x^{2} \, dx, \; w = x^{3}/3$

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$\int x^{2} \cdot ln(x) \, dx = \frac{x^{3}}{3}ln(x) - \frac{1}{3} \int x^{2} \, dx = \frac{x^{3}}{3}ln(x) - x^{3}/9 + C$

Substituting this into the first integral gets

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$G(x) = \frac{x^{3}}{3}[ \, ln(x)^{2} - \frac{2}{3}ln(x) + \frac{2}{9} \, ] + C$

$G(e) = \frac{5e^{3}}{27} + C = \frac{5e^{3}}{27} - 1, \; C = -1$

$G(x) = \frac{x^{3}}{3}[ \, ln(x)^{2} - \frac{2}{3}ln(x) + \frac{2}{9} \, ] - 1$