Question

Resolva a integral usando frações parciais
$\displaystyle \int \dfrac{\mathtt{x^{3}+x^{2}+2x+1}}{\mathtt{(x^{2}+1)(x^{2}+2)}} \,\mathtt{dx}$
Alguém pode me ajudar?

Answer 1

• A resposta dessa integral é 1/2 ln | x² + 1 | + 1/ √2 arctan( x/√2 ) + k.

Bom, queremos resolver a seguinte integral:

                       $\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}\end{gathered}$

A maneira mais fácil de resolver é utilizando frações parciais. Logo, temos que:

       $\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \int \left(\frac{x}{(x^2+1)} + \frac{1}{(x^2 +2)}\right)dx\end{gathered}$

Aplicando então a propriedade de integração.

    $\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \int \frac{x}{(x^2+1)} dx + \int \frac{1}{(x^2 +2)}dx\end{gathered}$

Resolvendo a primeira integral pelo método da substituição simples. Fazendo então:

$\large\displaystyle\begin{gathered} \int \frac{x}{(x^2+1)} dx \Rightarrow \begin{cases}u=x^2+1\\ du=2x dx\\ \frac{du}{2} = xdx \end{cases}\end{gathered}$

Substituindo.

   $\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \int \frac{x}{(x^2+1)} dx + \int \frac{1}{(x^2 +2)}dx\end{gathered}$

       $\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \int \frac{1}{u}\cdot \frac{du}{2} + \int \frac{1}{(x^2 +2)}dx\end{gathered}$

     $\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \frac{1}{2}\cdot \int \frac{du}{u} +\int \frac{1}{(x^2 +2)}dx\end{gathered}$

  $\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \frac{1}{2}\cdot \ln| x^2+1|+\int \frac{1}{(x^2 +2)}dx\end{gathered}$

Resolvendo agora a outra integral, também pelo método da substituição.

             $\large\displaystyle\text{ \begin{gathered} \int \frac{1}{(x^2+2)} dx \Rightarrow \int \frac{1}{2\left( \frac{x^2}{2 }+1 \right)}dx\end{gathered} }$

      $\large\displaystyle\text{ \begin{gathered} \frac{1}{2}\cdot \int \frac{1}{\left( \frac{x^2}{2 }+1 \right)}dx\Rightarrow \begin{cases} u^2= \frac{x^2}{2}\ ;\ u = \frac{x}{\sqrt{2} } \\\\ du= \frac{1}{\sqrt{2} } dx \end{cases}\end{gathered} }$

      $\large\displaystyle\text{ \begin{gathered} \frac{1}{2} \cdot \int \frac{1}{\left( \frac{x^2}{2 }+1 \right)}dx \Rightarrow \frac{1}{2} \cdot \int\frac{1}{u^2+ 1 } du\cdot \sqrt{2} \end{gathered} }$

     $\large\displaystyle\text{ \begin{gathered} \frac{1}{2} \cdot \int \frac{1}{\left( \frac{x^2}{2 }+1 \right)}dx \Rightarrow \frac{1}{2} \cdot \arctan(u)\cdot \sqrt{2} \end{gathered} }$

     $\large\displaystyle\text{ \begin{gathered} \frac{1}{2} \cdot \int \frac{1}{\left( \frac{x^2}{2 }+1 \right)}dx \Rightarrow \frac{1}{\sqrt{2} } \cdot \arctan\left(\frac{x}{\sqrt{2} } \right) + k \end{gathered} }$

Ficando então:

$\large\displaystyle\begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \frac{1}{2}\cdot \ln| x^2+1|+\int \frac{1}{(x^2 +2)}dx\end{gathered}$

$\large\displaystyle\text{ \begin{gathered} \int \frac{x^3 + x^2 + 2x + 1}{(x^2 +1) (x^2+2)}dx \Rightarrow \therefore \boxed{\boxed{\green{\frac{1}{2}\cdot \ln| x^2+1|+\frac{1}{\sqrt{2}}\arctan \left( \frac{x}{\sqrt{2} } \right) +k}}}\end{gathered} }$

Veja mais sobre:

Integrais indefinidas.

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Answer 2

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}\,dx\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}=\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+2}\bigg[(x^2+1)(x^2+2)\bigg]\\\\\sf x^3+x^2+2x+1=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)\\\sf x^3+x^2+2x+1=Ax^3+2Ax+Bx^2+2B+Cx^3+Cx+Dx^2+D\\\sf x^3+x^2+2x+1=(A+C)x^3+(B+D)x^2+(2A+C)x+2B+D\\\begin{cases}\sf A+C=1\\\sf B+D=1\\\sf 2A+C=2\\\sf 2B+D=1\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\begin{cases}\sf A+C=1\\\sf B+D=1\\\sf -C=0\\\sf-B=0\end{cases}\\\sf -B=0\implies B=0\\\sf -C=0\implies C=0\\\sf 0+D=1\implies D=1\\\sf A+0=1\implies A=1\end{array}}$$\boxed{\begin{array}{l}\sf \dfrac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}=\dfrac{x}{x^2+1}+\dfrac{1}{x^2+2}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}\,dx=\int\dfrac{x}{x^2+1}dx+\int\dfrac{dx}{x^2+2}\\\displaystyle\sf\int\dfrac{x}{x^2+1}dx=\dfrac{1}{2}\int\dfrac{2x}{x^2+1}dx\\\underline{\rm fac_{\!\!,}a}\\\sf t=x^2+1\implies dt=2x\,dx\\\displaystyle\sf\dfrac{1}{2}\int\dfrac{2x}{x^2+1}dx=\dfrac{1}{2}\int\dfrac{dt}{t}=\dfrac{1}{2}\ell n|t|+k\\\displaystyle\sf\int\dfrac{x}{x^2+1}\,dx=\dfrac{1}{2}\ell n|x^2+1|+k\\\displaystyle\sf\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}arctg\bigg(\dfrac{x}{a}\bigg)+k\\\displaystyle\sf\int\dfrac{dx}{x^2+2}=\int\dfrac{dx}{x^2+(\sqrt{2})^2}=\dfrac{1}{\sqrt{2}}arctg\bigg(\dfrac{x}{\sqrt{2}}\bigg)+k\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}\,dx=\dfrac{1}{2}\ell n|x^2+1|+\dfrac{1}{\sqrt{2}}arctg\bigg(\dfrac{x}{\sqrt{2}}\bigg)+k\end{array}}$