Ensino Superior ⇒ Integral por frações parciais Tópico resolvido

[julianonara]

[tex3]\int\limits_{1}^{2}\frac{dx}{x^{3}+6x^{2}+11x+6}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{1}^{2}\frac{dx}{x^{3}+6x^{2}+11x+6}[/tex3]

Resposta

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0,0266

[rcompany]

[tex3]f(x)=x^3+6x^2+11x+6\\
\text{Notemos que }f(-1)=0 \text{, então }\exists a,b\in\mathbb{R}/ f(x)=(x+1)(x^2+ax+b)\\
\begin{array}{rl}
x^3+6x^2+11x+6=(x+1)(x^2+ax+b)&\!\!\implies x^3+6x^2+11x+6=x^3+(a+1)x^2+(a+b)x+b\\
&\!\!\implies\left\{\begin{array}{l} a+1=6\\a+b=11\\b=6\end{array}\right.\\
&\!\!\implies\left\{\begin{array}{l} a=5\\b=6\end{array}\right.\\
&\!\!\implies f(x)=(x+1)(x^2+5x+6)\\
&\!\!\implies f(x)=(x+1)(x+2)(x+3)\quad\text{já que }\!-\!2\text{ e }\!-\!3\text{ são raízes de }x^2+5x+6

\end{array}\\[36pt]
\exists a,b,c\in\mathbb{R}/\forall x\in\mathbb{R},\dfrac{1}{(x+1)(x+2)(x+3)}=\dfrac{a}{x+1}+\dfrac{b}{x+2}+\dfrac{c}{x+3}\\
\begin{array}{rl}
\dfrac{1}{(x+1)(x\!+\!2)(x\!+\!3)}=\dfrac{a}{x\!+\!1}\!+\!\dfrac{b}{x\!+\!2}\!+\!\dfrac{c}{\!x+\!3}&\!\!\implies \dfrac{1}{(x+1)(x\!+\!2)(x\!+\!3)}=\dfrac{a(x\!+\!2)(x\!+\!3)\!+\!b(x\!+\!1)(x\!+\!3)\!+\!c(x\!+\!1)(x\!+\!2)}{(x\!+\!1)(x\!+\!2)(x\!+\!3)}\\
&\!\!\implies 0x^2+0x+1=(a+b+c)x^2+(5a+4b+3c)x+(6a+3b+2c)\\
&\!\!\implies\left\{\begin{array}{l}a+b+c=0\\
5a+4b+3c=0\\
6a+3b+2c=1\\
\end{array}\right.\\
&\!\!\implies\left\{\begin{array}{l}
a=\dfrac{1}{2}\\
b=-1\\
c=\dfrac{1}{2}
\end{array}\right.\\
&\implies \dfrac{1}{(x+1)(x+2)(x+3)}=\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}+\dfrac{1}{2(x+3}

\end{array}\\[24pt]
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]f(x)=x^3+6x^2+11x+6\\
\text{Notemos que }f(-1)=0 \text{, então }\exists a,b\in\mathbb{R}/ f(x)=(x+1)(x^2+ax+b)\\
\begin{array}{rl}
x^3+6x^2+11x+6=(x+1)(x^2+ax+b)&\!\!\implies x^3+6x^2+11x+6=x^3+(a+1)x^2+(a+b)x+b\\
&\!\!\implies\left\{\begin{array}{l} a+1=6\\a+b=11\\b=6\end{array}\right.\\
&\!\!\implies\left\{\begin{array}{l} a=5\\b=6\end{array}\right.\\
&\!\!\implies f(x)=(x+1)(x^2+5x+6)\\
&\!\!\implies f(x)=(x+1)(x+2)(x+3)\quad\text{já que }\!-\!2\text{ e }\!-\!3\text{ são raízes de }x^2+5x+6

\end{array}\\[36pt]
\exists a,b,c\in\mathbb{R}/\forall x\in\mathbb{R},\dfrac{1}{(x+1)(x+2)(x+3)}=\dfrac{a}{x+1}+\dfrac{b}{x+2}+\dfrac{c}{x+3}\\
\begin{array}{rl}
\dfrac{1}{(x+1)(x\!+\!2)(x\!+\!3)}=\dfrac{a}{x\!+\!1}\!+\!\dfrac{b}{x\!+\!2}\!+\!\dfrac{c}{\!x+\!3}&\!\!\implies \dfrac{1}{(x+1)(x\!+\!2)(x\!+\!3)}=\dfrac{a(x\!+\!2)(x\!+\!3)\!+\!b(x\!+\!1)(x\!+\!3)\!+\!c(x\!+\!1)(x\!+\!2)}{(x\!+\!1)(x\!+\!2)(x\!+\!3)}\\
&\!\!\implies 0x^2+0x+1=(a+b+c)x^2+(5a+4b+3c)x+(6a+3b+2c)\\
&\!\!\implies\left\{\begin{array}{l}a+b+c=0\\
5a+4b+3c=0\\
6a+3b+2c=1\\
\end{array}\right.\\
&\!\!\implies\left\{\begin{array}{l}
a=\dfrac{1}{2}\\
b=-1\\
c=\dfrac{1}{2}
\end{array}\right.\\
&\implies \dfrac{1}{(x+1)(x+2)(x+3)}=\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}+\dfrac{1}{2(x+3}

\end{array}\\[24pt]
[/tex3]

[tex3]
\text{E então:}\\
\begin{array}{rl}
\displaystyle\int_1^2\dfrac{\mathrm{d}x}{(x+1)(x+2)(x+3)}&=\displaystyle\int_1^2\dfrac{\mathrm{d}x}{2(x+1)}-\int_1^2\dfrac{\mathrm{d}x}{x+2}+\int_1^2\dfrac{\mathrm{d}x}{2(x+3)}\\
&=\displaystyle\int_1^2\dfrac{(\ln(x+1))^\prime}{2}\mathrm{d}x-\int_1^2(\ln(x+2))^\prime\mathrm{d}x+\int_1^2\dfrac{(\ln(x+3))^\prime}
{2}\mathrm{d}x\\
&=\left(\dfrac{\ln(3)}{2}-\dfrac{\ln(2)}{2}\right)-\Large\left(\normalsize\ln(4)-ln(3)\Large\right)\normalsize+\left(\dfrac{\ln(5)}{2}-\dfrac{\ln(4)}{2}\right)\\
&=\ln\left(\sqrt{\dfrac{3}{2}}\right)-\ln\left(\dfrac{4}{3}\right)+\ln\left(\sqrt{\dfrac{5}{4}}\right)\\
&=\ln\left(\dfrac{3}{4}\sqrt{\dfrac{3}{2}}\sqrt{\dfrac{5}{4}}\right)=\ln\left(\dfrac{3\sqrt{15}}{8\sqrt{2}}\right)\approx 0,0266
\end{array}



[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\text{E então:}\\
\begin{array}{rl}
\displaystyle\int_1^2\dfrac{\mathrm{d}x}{(x+1)(x+2)(x+3)}&=\displaystyle\int_1^2\dfrac{\mathrm{d}x}{2(x+1)}-\int_1^2\dfrac{\mathrm{d}x}{x+2}+\int_1^2\dfrac{\mathrm{d}x}{2(x+3)}\\
&=\displaystyle\int_1^2\dfrac{(\ln(x+1))^\prime}{2}\mathrm{d}x-\int_1^2(\ln(x+2))^\prime\mathrm{d}x+\int_1^2\dfrac{(\ln(x+3))^\prime}
{2}\mathrm{d}x\\
&=\left(\dfrac{\ln(3)}{2}-\dfrac{\ln(2)}{2}\right)-\Large\left(\normalsize\ln(4)-ln(3)\Large\right)\normalsize+\left(\dfrac{\ln(5)}{2}-\dfrac{\ln(4)}{2}\right)\\
&=\ln\left(\sqrt{\dfrac{3}{2}}\right)-\ln\left(\dfrac{4}{3}\right)+\ln\left(\sqrt{\dfrac{5}{4}}\right)\\
&=\ln\left(\dfrac{3}{4}\sqrt{\dfrac{3}{2}}\sqrt{\dfrac{5}{4}}\right)=\ln\left(\dfrac{3\sqrt{15}}{8\sqrt{2}}\right)\approx 0,0266
\end{array}



[/tex3]