Question

Integral definida:
$\int\limits^2_1 {x} (lnx) \, dx$

Answer 1

$\boxed{\boxed{\int\limits_{1}^{2}x\cdot ln(x)dx=\left\int x\cdot ln(x)dx\right]^{2}_{1}}}$

Então, a princípio, vamos trabalhar na integral indefinida:

$\int xln(x)dx\\\\\\u=ln(x)~~~~\rightarrow~~~~du=\frac{1}{x}dx\\v=\frac{x^{2}}{2}~~~~~~~~\leftarrow~~~~dv=xdx$

Integrando por partes:

$\int udv=uv-\int vdu\\\\\int xln(x)dx=ln(x)\cdot\dfrac{x^{2}}{2}-\int\dfrac{x^{2}}{2}\dfrac{1}{x}dx\\\\\int xln(x)dx=\dfrac{x^{2}ln(x)}{2}-\dfrac{1}{2}\int xdx\\\\\int xln(x)dx=\dfrac{x^{2}ln(x)}{2}-\dfrac{1}{2}\dfrac{x^{2}}{2}+C\\\\\int xln(x)dx=\dfrac{2x^{2}ln(x)}{4}-\dfrac{x^{2}}{4}+C\\\\\\\boxed{\boxed{\int xln(x)dx=\dfrac{x^{2}}{4}\left(2ln(x)-1\right)+C}}$
__________________________

$\int\limits_{1}^{2}xln(x)dx=\left[\dfrac{x^{2}}{4}\left(2ln(x)-1\right)+C\right]^{2}_{1}\\\\\\\int\limits_{1}^{2}xln(x)dx=\left(\dfrac{2^{2}}{4}(2ln(2)-1)\right)-\left(\dfrac{1^{2}}{4}\left(2ln(1)-1\right)\right)\\\\\\\int\limits_{1}^{2}xln(x)dx=1(2\cdot ln(2)-1)-\dfrac{1}{4}(2\cdot0-1)\\\\\\\int\limits_{1}^{2}xln(x)dx=ln(2^{2})-1-\dfrac{1}{4}(-1)\\\\\\\int\limits_{1}^{2}xln(x)dx=ln(4)-1+\dfrac{1}{4}\\\\\\\boxed{\boxed{\int\limits_{1}^{2}xln(x)dx=ln(4)-\dfrac{3}{4}}}$