Question

Resolva a integral:

$\large\displaystyle\begin{aligned}\int \dfrac{4x^2+2x+4}{x^3+4x}\end{aligned}$

Answer 1

$\boxed{\begin{array}{l}\sf\dfrac{4x^2+2x+4}{x^3+4x}=\dfrac{4x^2+2x+4}{x\cdot(x^2+4)}\\\\\sf\dfrac{4x^2+2x+4}{x\cdot(x^2+4)}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+4}\cdot[x\cdot(x^2+4)]\\\\\sf 4x^2+2x+4=A\cdot(x^2+4)+(Bx+C)\cdot x\\\sf 4x^2+2x+4=Ax^2+4A+Bx^2+CX\\\sf 4x^2+2x+4=(A+B)x^2+Cx+4A\\\sf pela~identidade~de~polin\hat omios~podemos~escrever:\\\begin{cases}\sf A+B=4\\\sf C=2\\\sf 4A=4\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\sf4A=4\\\sf A=\dfrac{4}{4}\\\sf A=1\\\sf A+B=4\\\sf 1+B=4\\\sf B=4-1\\\sf B=3\\\sf portanto\\\sf\dfrac{4x^2+2x+4}{x\cdot(x^2+4)}=\dfrac{1}{x}+\dfrac{3x+2}{x^2+4}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{4x^2+2x+4}{x^3+4x}dx=\int\dfrac{4x^2+2x+4}{x\cdot(x^2+4)}=\int\dfrac{1}{x}dx+\int\dfrac{3x+2}{x^2+4}dx\\\displaystyle\sf\int\dfrac{4x^2+2x+4}{x^3+4x}dx=\int\dfrac{1}{x}dx+\dfrac{3}{2}\int\dfrac{2x}{x^2+4}dx+2\int\dfrac{1}{x^2+4}dx\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\dfrac{3}{2}\int\dfrac{2x}{x^2+4}dx\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf t=x^2+4\implies dt=2xdx\\\displaystyle\sf\dfrac{3}{2}\int\dfrac{2x}{x^2+4}dx=\dfrac{3}{2}\int\dfrac{dt}{t}=\dfrac{3}{2}\ell n |t|+c_1\\\displaystyle\sf\dfrac{3}{2}\int\dfrac{2x}{x^2+4}dx=\dfrac{3}{2}\ell n|x^2+4|+c_1\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{dt}{a^2+t^2}=\dfrac{1}{a}arctg\bigg(\dfrac{t}{a}\bigg)+c\\\displaystyle\sf\int\dfrac{1}{x^2+4}dx=\int\dfrac{1}{x^2+2^2}dx=\dfrac{1}{2}arctg\bigg(\dfrac{x}{2}\bigg)+c_2\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{dx}{x}=\ell n|x|+c_3\end{array}}$

$\large\boxed{\begin{array}{l}\sf substituindo~os~resultados~obtidos~temos:\\\displaystyle\sf\int\dfrac{4x^2+2x+4}{x^3+4x}dx=\ell n|x|+\dfrac{3}{2}\ell n |x^2+4|+\diagup\!\!\!2\cdot\dfrac{1}{\diagup\!\!\!2}arctg\bigg(\dfrac{x}{2}\bigg)+k\\\rm podemos~simplificar~ainda~mais~veja:\\\displaystyle\sf\int\dfrac{4x^2+2x+4}{x^3+4x}dx=\ell n\bigg|x\cdot(x^2+4)^{\frac{3}{2}}\bigg|+arctg\bigg(\dfrac{x}{2}\bigg)+k\end{array}}$

Answer 2

Resposta para a integral:

• ln lxl + 3/2 ln l x² + 4 l + arctg (x/2) + C

Integral - Frações parciais

Temos a Seguinte integral:

$\Large \displaystyle \int \sf \dfrac{4{x}^{2} + 2x + 4 }{ {x}^{3} + x }$

Resolvendo a integral pelo método de frações parciais. Vamos primeiramente fatorar o denominador e igualar aos termos A, B e C

$\Large \boxed{ \begin{array}{c} \\ \sf \dfrac{4 {x}^{2} + 2x + 4}{ {x}^{3} + 4x} = \dfrac{4 {x}^{2} + 2x + 4}{ x({x}^{2} + 4)} \\ \\ \sf\dfrac{4 {x}^{2} + 2x + 4}{ x({x}^{2} + 4)} = \dfrac{A}{x} + \dfrac{Bx + C}{ {x}^{2} + 4 } \\ \: \end{array}}$

• Perceba que, no caso da integral, temos o denominador x(x²+3), se tentarmos simplificar mais, teríamos > x = -3, e daria raiz com número negativo, portanto, esse polinômio é considerado irredutível

Agora temos que achar o valor de A, B e C, dividindo denominador por x(x²+4) e Multiplicando o resultado pelas letras, teremos:

$\Large \boxed{\begin{array}{c} \sf \dfrac{4x^2 +2x+4}{\cancel{x(x^2+4)}} = \dfrac{A(x^2 + 4 ) + ( Bx + C )x}{ \cancel{ x(x^2 + 4)} } \\\\\sf 4x^2 +2x+4 = Ax^2 + 4A+ Bx^2 + Cx \\\\\sf 4x^2 +2x+4 = (A+B)x^2 + Cx + 4A \\\: \end{array}} </p><p></p><p>Fazendo por <strong>igualdade </strong><strong>de </strong><strong>polinômios,</strong><strong> </strong>(A+B) = 4<strong>,</strong><strong> </strong><strong>(</strong><strong> </strong><strong>x²</strong><strong> </strong><strong>)</strong><strong>,</strong><strong> </strong>C = 2, <strong>(</strong><strong> </strong><strong>x </strong><strong>)</strong><strong> </strong>e 4A = 4 <strong>(</strong><strong> </strong><strong>termo </strong><strong>independente </strong><strong>)</strong><strong>.</strong><strong> </strong>Veja o cálculo Abaixo:</p><p></p><p>[tex]\Large \begin{cases} \sf A + B = 4 \Rightarrow B = 4-1 \: \: \Lambda \: \: B = 3 </p><p></p><p> \\ \sf C = 2 \\\sf 4A = 4 \Rightarrow A = \dfrac{4}{4} \: \: \Lambda \: \: \: A = 1 </p><p></p><p>\end{cases}$

• Definido os valores, Temos que:

$\huge \sf \dfrac{1}{x} + \dfrac{3x + 2}{ {x}^{2} + 4}$

Resolvendo as integrais separadamente, vamos primeiro resolver 1/x depois 3x+2/x²+ 4.

• Na integral de 1/x nem precisa de Cálculo, é só aplicar a regra:

$\Large \boxed{ \boxed{ \displaystyle\int \sf \frac{1}{x} dx = ln \: lxl + C}}$

Resolvendo a integral 3x + 2/x² + 4, resolvendo por partes, vamos fazer 3x e depois 2, Cálculo das integrais abaixo:

$\Large \boxed{\begin{array}{c} \displaystyle\int \sf \frac{3x}{x^2+4} dx \\\\\sf U = x^2 - 4 \rightarrow U = 2x \\ \sf \dfrac{du}{dx} = 2xdx \rightarrow \dfrac{1}{2} du = xdx \\ \\ \sf logo > 3\cdot\frac{1}{2} \displaystyle\int \sf\dfrac{1}{U} du = \dfrac{3}{2} ln lUl + C \\\\\sf = \dfrac{3}{2} ln \: lx^2+4l + C \\\: \end{array}}$

Resolvendo a outra integral

$\Large \boxed{\begin{array}{c}\\ \displaystyle\int \sf \frac{2}{x^2+4} dx \Rightarrow \sf 2\displaystyle\int \sf \frac{2}{4+x^2} dx \\\\\sf 2 \displaystyle\int \sf \frac{1}{4\Bigg( 1+ \dfrac{x^2}{4}\Bigg)} \Rightarrow \sf \dfrac{2^{\div2}}{4^{\div2}} \displaystyle\int \sf 1+\frac{1}{ 1+\Bigg( \dfrac{x}{2} \Bigg) ^{2} } dx \\ \: \end{array}}$

Seja U = x/2, então du = 1/2 dx > dx = 2du

• Lembrando da regra:

$\huge \sf \int \sf \dfrac{1}{1+u^2} du = arctg u + C$

Logo:

$\Large \boxed{\begin{array}{c} \\\sf \dfrac{1}{2} \displaystyle\int \sf \dfrac{1}{1+u^2} 2du \Rightarrow \dfrac{2}{2} \displaystyle\int \sf \dfrac{1}{1+u^2} du \\\\\sf = 1. arctg u + C \\\\\sf arctg \Bigg(\dfrac{x}{2} \Bigg)+C \\\: \end{array}}$

Juntando os resultados, temos como resposta:

$\Large \boxed{\boxed{\sf ln \: lxl + \dfrac{3}{2} ln \: l x^2 + 4 l + arctg \Bigg(\dfrac{x}{2} \Bigg)+ C }}$

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