Matemática, perguntado por FlaviaAlves, 11 meses atrás

integral raiz quadrada x lnx dx

Soluções para a tarefa

Respondido por albertrieben
2

Integral √x * ln(x) dx

Para o integrando √x ln(x) dx ,

integre por partes, ∫ f dg = f g - ∫g df, em que

 f = log(x), dg = √(x) dx, df = 1/x dx, g = (2 x^(3/2))/3:

A integral de √(x) é (2 x ^ (3/2)) / 3:

 = 2/3 x ^ (3/2) ln(x) - (4 x ^ (3/2)) / 9 + C

Resposta:

 | = 2/9 x ^ (3/2) (3 ln(x) - 2) + C

Respondido por solkarped
5

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

         

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Solução gráfica (figura):

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