Question

Use a técnica de integração adequada para
calcular:

Answer 1

• Resolvendo essas integrais indefinidas pelos métodos adequados, temos que:

$\Large\displaystyle\begin{gathered} \sf a)\int \frac{3x+4}{x^2-5x+6} dx=-10\ln|x-2|+13\ln|x-3|+\mathbb{C}\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \sf b)\int x^2e^{3x}dx= \frac{e^{3x}x^2}{3} -\frac{2e^{3x}x}{9}+\frac{2e^{3x}}{27} +\mathbb{C} \end{gathered} }$

Desejamos calcular as seguintes integrais:

$\Large\displaystyle\begin{gathered} \sf \int \frac{3x+4}{x^2-5x+6} dx\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf \int x^2e^{3x}dx\end{gathered}$

Na primeira integral, devemos aplicar o método das frações parciais, ficando então:

$\Large\displaystyle\begin{gathered} \sf \int \frac{3x+4}{x^2-5x+6} dx = \int \frac{3x+4}{(x-2)(x-3)}dx\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf \frac{3x+4}{(x-2)(x-3)}=\frac{A}{(x-2)} +\frac{B}{(x-3)} \end{gathered}$ $\Large\displaystyle\begin{gathered} \sf \frac{3x+4}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)} \end{gathered}$

Agora, perceba que como temos uma igualdade e os denominadores são iguais, logo, os numeradores também são.

$\Large\displaystyle\begin{gathered} \sf A(x-3)+B(x-2)=3x+4\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf Ax-3A+Bx-2B-3x-4=0\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf x(A+B-3)+(-3A-2B-4)=0\end{gathered}$

E com isso, temos o seguinte sistema:

$\Large\displaystyle\begin{gathered} \begin{cases}\sf A+B=3\ \ \ \green{(i)}\\ \sf -3A -2B=4\ \ \ \green{(ii)}\end{cases}\end{gathered}$

Resolvendo:

$\Large\displaystyle\begin{gathered} \sf A=3-B\ \ \ \green{(iii)}\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf -3(3-B)-2B=4\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf -9+3B-2B=4\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\therefore \underline{\boxed{\sf B=13}}\end{gathered} }$

$\Large\displaystyle\begin{gathered} \sf A=3-B\ \ \ \green{(iii)}\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf A=3-13\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \therefore \underline{\boxed{\sf A=-10}}\end{gathered} }$

Substituindo os valores de A e B:

$\Large\displaystyle\begin{gathered} \sf \int\frac{3x+4}{(x-2)(x-3)} dx=\int-\frac{10}{x-2} dx +\int\frac{13}{x-3}dx \end{gathered}$

E pela lineariedade, temos que:

$\Large\displaystyle\begin{gathered} \sf \int\frac{3x+4}{(x-2)(x-3)} dx=-10\int\frac{1}{x-2} dx +13\int\frac{1}{x-3}dx \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \sf \int\frac{3x+4}{(x-2)(x-3)} dx=\green{\underline{\boxed{\sf -10\ln|x-2| +13\ln|x-3| +\mathbb{C}}}}\end{gathered} }$

Agora, vamos resolver a outra integral, mas dessa vez, perceba que estamos integrando um produto de funções, logo, aplicaremos o método da integração por partes. Chamando então:

$\Large\displaystyle\begin{gathered} \sf u=x^2 \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf du= 2xdx \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf dv=e^{3x}dx\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf v=\frac{1}{3}e^{3x}\end{gathered}$

• Logo:

$\Large\displaystyle\text{ \begin{gathered} \sf \int x^2e^{3x}dx=\frac{e^{3x}x^2}{3}-\int\frac{2}{3}e^{3x}xdx\end{gathered} }$

Pela lineariedade:

$\Large\displaystyle\text{ \begin{gathered} \sf \int x^2e^{3x}dx=\frac{e^{3x}x^2}{3}-\frac{2}{3}\underbrace{\int e^{3x}xdx}_{P/partes\ dnv}\end{gathered} }$

Chamando então:

$\Large\displaystyle\begin{gathered} \sf u=x \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf du=dx \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf dv=e^{3x}dx \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf v=\frac{1}{3}e^{3x}\end{gathered}$

• Logo;

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

$\Large\displaystyle\text{ \begin{gathered} \sf \int x^2e^{3x}dx=\frac{e^{3x}x^2}{3}-\frac{2}{3}\left[\frac{e^{3x}x}{3}-\frac{e^{3x}}{9}\right]\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \sf \int x^2e^{3x}dx=\green{\underline{\boxed{\sf\frac{e^{3x}x^2}{3}-\frac{2e^{3x}x}{9}+\frac{2e^{3x}}{27} +\mathbb{C}}}}\end{gathered} }$

 

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