Ohne Wolfi und alles schön zu Fuss:
$$ \int x \cdot (\ln (x))^2 \, dx $$
$$ \frac{ d\, \ln(x)^2}{dx}=\frac 1x \,\cdot \, 2 \ln (x) $$
$$ \int x \cdot (\ln (x))^2 \, dx = \frac12 x^2 \cdot \, (\ln (x))^2 - \int \, \frac12 x^2 \cdot \,\frac 1x \,\cdot \, 2 \ln (x) dx$$
$$ \int x \cdot (\ln (x))^2 \, dx = \frac12 x^2 \cdot \, (\ln (x))^2 - \int \, x \,\cdot \, \ln (x) dx$$
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$$ \int \, x \,\cdot \, \ln (x) dx = \frac12 x^2 \cdot \, \ln (x) - \int \, \frac12 x^2 \cdot \,\frac 1x dx$$
$$ \int \, x \,\cdot \, \ln (x) dx = \frac12 x^2 \cdot \, \ln (x) - \int \, \frac12 x \,\, dx$$
$$ \int \, x \,\cdot \, \ln (x) dx = \frac12 x^2 \cdot \, \ln (x) - \frac12 \cdot \frac12 x^2 \,\, $$
$$ \int \, x \,\cdot \, \ln (x) dx = \frac12 x^2 \cdot \,\left( \ln (x) - \frac12 \right) \,\, $$
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$$ \int x \cdot (\ln (x))^2 \, dx = \frac12 x^2 \cdot \, (\ln (x))^2 - \frac12 x^2 \cdot \,\left( \ln (x) - \frac12 \right)$$
$$ \int x \cdot (\ln (x))^2 \, dx = \frac12 x^2 \cdot \,\left( (\ln (x))^2 - \ln (x) - \frac12 \right)$$
$$ \int x \cdot (\ln (x))^2 \, dx = \frac{x^2}4 \cdot \,\left( 2\,(\ln (x))^2 - 2 \, \ln (x) -1 \right) +C $$