Matemática, perguntado por gideaoengenharia, 1 ano atrás

integral x^1/2 lnx dx

Soluções para a tarefa

Respondido por Usuário anônimo

∫  √x * ln x dx

Por partes :

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

√x dx = dv ==> ∫ √x dx = ∫ dv => (2/3) * x^(3/2) = v

∫  √x * ln x dx = (2/3) * x^(3/2) * ln x - ∫   (2/3) * x^(3/2) * (1/x) dx

∫  √x * ln x dx = (2/3) * x^(3/2) * ln x -(2/3)* ∫x^(3/2-1)) dx

∫  √x * ln x dx = (2/3) * x^(3/2) * ln x -(2/3)* ∫x^(1/2) dx

∫  √x * ln x dx = (2/3) * x^(3/2) * ln x -(2/3)* ∫√x dx

∫  √x * ln x dx = (2/3) * x^(3/2) * ln x -(2/3)* (2/3)x^(3/2) + const

∫  √x * ln x dx = (2/3) * x^(3/2) * ln x -(4/9)* x^(3/2) + const

∫  √x * ln x dx =x^(3/2)* [(2/3)* ln x -(4/9)] + const

∫  √x * ln x dx =(2/9)*x^(3/2)* [3 ln x -2] + const

∫  √x * ln x dx =(2/9)* √x³ * [3 ln x -2] + const

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 x^{\frac{1}{2}}\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 x^{\frac{1}{2}}\cdot\ln x\,dx\end{gathered}$}$

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

Nomear as funções.

$\Large\begin{cases}\tt f(x) = \ln x\\\tt g(x) = x^{\frac{1}{2}}\end{cases}$

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 x^{\frac{1}{2}}\,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 x^{\frac{1}{2}}\ln x\,dx= \frac{2}{3}x^{\frac{3}{2}}\ln x - \frac{4}{9}x^{\frac{3}{2}} + c\end{gathered}$}$