Matemática, perguntado por douglastuka, 1 ano atrás

integracao de raiz x ln (x)dx

Soluções para a tarefa

Respondido por Lukyo

$I=\displaystyle\int{\sqrt{x}\,\mathrm{\ell n\,}x\,dx}\\ \\ \\ =\int{x^{1/2}\,\mathrm{\ell n\,}x\,dx}$

$\bullet\;\;$ Utilizaremos o método de integração por partes:

$\begin{array}{ll} u=\mathrm{\ell n\,}x\;\;&\;\;du=\dfrac{1}{x}\,dx\\ \\ dv=x^{1/2}\,dx\;\;&\;\;v=\dfrac{2}{3}\,x^{3/2}\\ \end{array}$

Substituindo na fórmula de integração por partes, temos

$\displaystyle\int{u\,dv}=uv-\int{v\,du}\\ \\ \\ \int{x^{1/2}\,\mathrm{\ell n\,}x\,dx}=\dfrac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\int{\dfrac{2}{3}\,x^{3/2}\cdot \dfrac{1}{x}\,dx}\\ \\ \\ I=\dfrac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\dfrac{2}{3}\int{x^{3/2}\cdot x^{-1}\,dx}\\ \\ \\ I=\dfrac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\dfrac{2}{3}\int{x^{1/2}\,dx}\\ \\ \\ I=\dfrac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\dfrac{2}{3}\cdot \left(\dfrac{2}{3}\,x^{3/2} \right )+C\\ \\ \\ I=\dfrac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\dfrac{4}{9}\,x^{3/2}+C\\ \\ \\ \boxed{\begin{array}{ccc}\\&\displaystyle\int{\sqrt{x}\,\mathrm{\ell n\,}x\,dx}=\dfrac{2}{3}\,x^{3/2}\,\mathrm{\ell n\,}x-\dfrac{4}{9}\,x^{3/2}+C&\\ \\ \end{array}}$

Respondido por solkarped

✅ Após resolver os cálculos, concluímos que a integral indefinida - primitiva ou antiderivada - procurada é:

$\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf \int \sqrt{x}\ln x\,dx= \frac{2}{3}x^{\frac{3}{2}}\ln x - \frac{4}{9}x^{\frac{3}{2}} + c\:\:\:}}\end{gathered}$}$

Seja a integral:

$\Large\displaystyle\text{$\begin{gathered}\tt \int \sqrt{x}\cdot\ln x\,dx\end{gathered}$}$

Para resolver esta questão devemos realizar a integração por partes. Para isso, devemos:

Nomear as funções.

$\Large\begin{cases}\tt f(x) = \ln x\\\tt g(x) = \sqrt{x}\end{cases}$

Desenvolver e simplificar os cálculos.

$\Large\displaystyle\text{$\begin{gathered}\tt I = \int \left[f(x)\cdot g(x)\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = f(x)\cdot\int g(x)\,dx - \int \left[\frac{d}{dx} f(x)\cdot\int g(x)\,dx\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \ln x\cdot\int \sqrt{x}\,dx - \int \left[\frac{d}{dx}\ln x\cdot\int \sqrt{x}\,dx\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \ln x\cdot\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \int \left[\frac{1}{x}\cdot\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \ln x\cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \int \left[\frac{1}{x}\cdot\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \frac{1}{\frac{3}{2}}\cdot x^{\frac{3}{2}}\ln x - \int \left[\frac{1}{\frac{3}{2}}\cdot\frac{x^{\frac{3}{2}}}{x}\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \frac{2}{3}x^{\frac{3}{2}}\ln x - \int \left[\frac{2}{3}\cdot x^{\frac{1}{2}}\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \frac{2}{3}x^{\frac{3}{2}}\ln x - \frac{2}{3}\cdot\frac{x^{\frac{1}{2}+ 1}}{\frac{1}{2} + 1} + c\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \frac{2}{3}x^{\frac{3}{2}}\ln x - \frac{2}{3}\cdot\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \frac{2}{3}x^{\frac{3}{2}}\ln x - \frac{2}{3}\cdot\frac{2}{3}\cdot x^{\frac{3}{2}} + c\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \frac{2}{3}x^{\frac{3}{2}}\ln x - \frac{4}{9}x^{\frac{3}{2}} + c\end{gathered}$}$

✅ Portanto, a integral procurada é:

$\Large\displaystyle\text{$\begin{gathered}\tt \int \sqrt{x}\ln x\,dx= \frac{2}{3}x^{\frac{3}{2}}\ln x - \frac{4}{9}x^{\frac{3}{2}} + c\end{gathered}$}$