<s>[INST] Calculate $\int^4_2 xln(x) \:dx$ using integration by parts. Let $f'(x) = x$ and $g(x) = \ln(x) \implies f(x) = \dfrac{x^2}{2}$ and $g'(x) =  \dfrac{1}{x}$.
$$\int^4_2 x\ln(x) \:dx = \left[ \dfrac{x^2}{2}\ln(x) \right]^4_2 - \int^4_2 \dfrac{1}{x}\dfrac{x^2}{2} \: dx = 8\ln(4) - 2\ln(2) - 1$$
Apparently, this answer is incorrect. I would greatly appreciate it if people could please explain what I'm doing wrong.
 [/INST] $$\int _{ 2 }^{ 4 } xln(x)\: dx=\left[ { \frac { x^{ 2 } }{ 2 }  }ln(x) \right] ^{ 4 }_{ 2 }-\int _{ 2 }^{ 4 }{ \frac { { x } }{ 2 }  } \: dx=\left[ { \frac { x^{ 2 } }{ 2 }  }ln(x) \right] ^{ 4 }_{ 2 }-\left[ { \frac { x^{ 2 } }{ 4 }  } \right] ^{ 4 }_{ 2 }=\frac { 16 }{ 2 } \ln { 4 } -\frac { { 2 }^{ 2 } }{ 2 } \ln { 2 } -\left[ \frac { 16 }{ 4 } -\frac { 4 }{ 4 }  \right] =\\ =14\ln { 2 } -3$$
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