Question

2) Resolva a integral e diga o método que está usando:

$\int\limits x^{2}~ln~x } \, dx$

Answer 1

Temos o seguinte:

$\int {x^2ln(x)} \, dx$

Usando integral por partes:

$\int\ {udv} \, = u\cdot v - \int\ {vdu} \,$

Temos:

$u = ln(x) \to \frac{du}{dx} = \frac{1}{x} \to du = \frac{1}{x} dx$

$dv = x^2 dx \to v = \int\ {x^2} \, dx \to v = \frac{x^3}{3}$

Logo temos:

$\int {x^2ln(x)} \, dx = ln(x) \cdot \frac{x^3}{3} - \int { \frac{x^3}{3} \cdot \frac{1}{x} } \, dx \\ \\ \int {x^2ln(x)} \, dx = \frac{ln(x) \cdot x^3}{3} - \int { \frac{x^2}{3} } \, dx \\ \\ \int {x^2ln(x)} \, dx = \frac{ln(x) \cdot x^3}{3} - \frac{1}{3} \int {x^2} \, dx \\ \\ \int {x^2ln(x)} \, dx = \frac{ln(x) \cdot x^3}{3} - \frac{x^3}{9} + C$

Answer 2

$\displaystyle\mathsf{\int(x^2ln(x))dx}$

$\mathsf{u=ln(x)\to~du=\dfrac{1}{x}dx}\\\mathsf{dv=x^2dx\to~v=\dfrac{1}{3}x^3}$

$\displaystyle\mathsf{\int(x^2ln(x)dx} \\\mathsf{=\dfrac{1}{3}x^3ln(x)-\dfrac{1}{3}\int(x^3.\dfrac{1}{x})dx}\\\displaystyle\mathsf{ \int \: x^2ln(x)dx = \dfrac{1}{3} {x}^{3} ln(x)-\dfrac{1}{3}\int~x^2\,dx}$

$\displaystyle\mathsf{\int~x^2ln(x)dx=\dfrac{1}{3}x^3-\dfrac{1}{9}x^3+k}$

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$\large\displaystyle\mathsf{Life=\int\limits_{birth}^{death}\dfrac{happiness}{time}dtime}$