Question

resolva integral de \int\ , x.lnx dx

Answer 1

Resolução da questão, veja:

Para essa integral, vamos integrar por partes:

∫ u dv = uv - ∫ v du

u = ln(x) => du = 1/x dx;

v = (x²/2) => dv = x dx

Veja:

∫ x ln(x) dx

½ x² ln(x) - ½ ∫ x dx

½ x² ln(x) - (x²/4)

Simplificando, teremos:

¼ x²(2 ln(x) - 1) + C

Ou seja, a integral acima resulta em:

¼ x²(2 ln(x) - 1) + C

Espero que te ajude. :-)

Answer 2

$\displaystyle\sf\int x\cdot\ell nx~dx\\\sf fac_{\!\!,}a~u=\ell nx\implies du=\dfrac{dx}{x}\\\sf dv=x~dx\implies v=\dfrac{1}{2}x^2\\\displaystyle\sf \int x\cdot \ell nx~dx=\dfrac{1}{2}x^2\cdot\ell nx-\int\dfrac{1}{2}\diagup\!\!\!x^2^\cdot\dfrac{1}{\diagup\!\!\!x}~dx\\\displaystyle\sf\int x\cdot\ell nx~dx=\dfrac{1}{2}x^2\cdot\ell nx-\dfrac{1}{2}\int x~dx\\\displaystyle\sf\int x\cdot\ell nx~dx=\dfrac{1}{2}x^2\ell nx=\dfrac{1}{2}\cdot\dfrac{1}{2}x^2+k$

$\displaystyle\sf\int x\cdot\ell nx~dx=\dfrac{1}{2}x^2\cdot\ell nx-\dfrac{1}{4}x^2+k$