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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.

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    $\begingroup$ What is the integral of $\frac{x}{2}$? $\endgroup$ – kingW3 Mar 5 '17 at 18:37
  • $\begingroup$ @kingW3 $\dfrac{x^2}{4}$? $\endgroup$ – The Pointer Mar 5 '17 at 18:39
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    $\begingroup$ $\frac{4^2}{4}-\frac{2^2}{4}$ is equal to? $\endgroup$ – kingW3 Mar 5 '17 at 18:39
  • $\begingroup$ @kingW3 You're absolutely correct. I apologise for the stupid error. Thank you very much for the assistance. $\endgroup$ – The Pointer Mar 5 '17 at 18:43
  • $\begingroup$ No problem,no need to apologize it happens to everybody,glad I could help. $\endgroup$ – kingW3 Mar 5 '17 at 18:44
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$$\int _{ 2 }^{ 4 } xln(x)\: dx=\left[ { \frac { x^{ 2 } }{ 2 } }ln(x) \right] ^{ 4 }_{ 2 }-\int _{ 2 }^{ 4 }{ \frac { { x } }{ 2 } } \: dx=\left[ { \frac { x^{ 2 } }{ 2 } }ln(x) \right] ^{ 4 }_{ 2 }-\left[ { \frac { x^{ 2 } }{ 4 } } \right] ^{ 4 }_{ 2 }=\frac { 16 }{ 2 } \ln { 4 } -\frac { { 2 }^{ 2 } }{ 2 } \ln { 2 } -\left[ \frac { 16 }{ 4 } -\frac { 4 }{ 4 } \right] =\\ =14\ln { 2 } -3$$

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You can also use a property of $\ln(x)$.

$$\int^4_2 x\ln(x) \:dx = \frac12\int^4_2 \ln(x) \:d(x^2) = \frac14\int^4_2 \ln(x^2) \:d(x^2) = \frac14\int^{16}_{4} \ln(x) \:d(x) = \frac14\left(x\ln(x) - x\right)\bigg\vert_{4}^{16} = 4\ln(16) - 4 - \ln(4) + 1 = 7\ln(4) - 3 $$

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$\int^4_2 xln(x) dx = \left[ \dfrac{x^2}{2}ln(x) \right]^4_2 - \int^4_2 \dfrac{1}{x}\dfrac{x^2}{2} \: dx = 8ln(4) - 2ln(2) - [\frac{x^2}{4}]^{4}_{2}=8ln (4)-2ln(2)-3$

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