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bom dia,
quero resolver esta integral de fração parcial que não consigo achar uma solução:

$\int_{}^{}\frac{x^{2}-3x+5}{3x^{3}+x^{2}+x} dx$

Quero a atenção de todos!
Obrigado.

$\\ \frac{x^2-3x+5}{3x^3+x^2+x}\\ \\ \frac{x^2-3x+5}{x.(3x^2+x+1)}\\ \\ \frac{A}{x}+\frac{Bx+C}{3x^2+x+1}\\ \\ A(3x^2+x+1)+Bx^2+Cx=x^2-3x+5\\ \\ 1.A=5\;\rightarrow\;A=5\\ \\ 3.A+B=1\;\rightarrow\;B=1-15=-14\\ \\ 1.A+1.C=-3\;\rightarrow\;C=-3-5=-8$

A integral fica:
$\int\;\left(\frac{5}{x}-\frac{14x+8}{3x^2+x+1}\right)dx$

O termo $\frac{5}{x}$ da integral é simples de resolver, vamos então nos concentrar no segundo termo:
$\int\;-\frac{14x+8}{3x^2+x+1}dx\\ \\ -2\int\;\frac{7x+4}{3x^2+x+1}dx\\ \\ Completando\;quadrado\;no\;denominador:\\ \\ 3x^2+x+1\;\;=\;\;3\left(x+\frac{1}{6}\right)^2+\frac{11}{12}\\ \\ -2\int\;\frac{7x+4}{3\left(x+\frac{1}{6}\right)^2+\frac{11}{12}}dx\\ \\ -\frac{2}{3}\int\;\frac{7x+4}{\left(x+\frac{1}{6}\right)^2+\frac{11}{36}}dx\\ \\ Utilizando \;o \;metodo\; da\; substituicao:\\ \\ u=x+\frac{1}{6}\\ du=dx\\ \\ -\frac{2}{3}\int\;\frac{7u+\frac{17}{6}}{\left(u\right)^2+\frac{11}{36}}dx\\ \\ -\frac{2}{3}\left(\int\;\frac{7u}{\left(u\right)^2+\frac{11}{36}}dx+\int\;\frac{\frac{17}{6}}{\left(u\right)^2+\frac{11}{36}}dx\right)$

$-\frac{2}{3}\left(\int\;\frac{7u}{\left(u\right)^2+\frac{11}{36}}dx+\int\;\frac{\frac{17}{6}}{\left(u\right)^2+\frac{11}{36}}dx\right)\\ \\ pela\;tabela\;de\;integrais:\\ \int\;\frac{\frac{17}{6}}{\left(u\right)^2+\frac{11}{36}}dx=\frac{17}{6}.\frac{1}{\sqrt{\frac{11}{36}}}.arctg\left(\frac{u}{\sqrt{\frac{11}{36}}}\right)=\frac{17}{\sqrt{11}}.arctg\left(\frac{u}{\sqrt{\frac{11}{36}}}\right)\\ \\ \\Por \;substituicao:\\\\ \int\;\frac{7u}{\left(u\right)^2+\frac{11}{36}}dx=\frac{7}{2}\int\;\frac{2dx}{\left(u\right)^2+\frac{11}{36}}=\frac{7}{2}ln\left|u^2+\frac{11}{46}\right|$

$-\frac{2}{3}\left(\int\;\frac{7u}{\left(u\right)^2+\frac{11}{36}}dx+\int\;\frac{\frac{17}{6}}{\left(u\right)^2+\frac{11}{36}}dx\right)\\ \\ pela\;tabela\;de\;integrais:\\ \int\;\frac{\frac{17}{6}}{\left(u\right)^2+\frac{11}{36}}dx=\frac{17}{6}.\frac{1}{\sqrt{\frac{11}{36}}}.arctg\left(\frac{u}{\sqrt{\frac{11}{36}}}\right)=\frac{17}{\sqrt{11}}.arctg\left(\frac{u}{\sqrt{\frac{11}{36}}}\right)\\ \\ \\Por \;substituicao:\\\\ \int\;\frac{7u}{\left(u\right)^2+\frac{11}{36}}dx=\frac{7}{2}\int\;\frac{2dx}{\left(u\right)^2+\frac{11}{36}}=\frac{7}{2}ln\left|u^2+\frac{11}{46}\right| \\\\ Juntando\; tudo,\; temos:\\ \\ \int\;\frac{5}{x}dx-\frac{2}{3}\left(\frac{17}{\sqrt{11}}.arctg\left(\frac{u}{\sqrt{\frac{11}{36}}}\right)\;+\;\frac{7}{2}ln\left|u^2+\frac{11}{46}\right|\right)\\ \\ 5.ln|x|-\frac{34}{3\sqrt{11}}.arctg\left(\frac{u}{\sqrt{\frac{11}{36}}}\right)-\frac{21}{4}ln\left|u^2+\frac{11}{46}\right|\right) \\ \\ Voltando \;a \;substituicao\; de\; u: \\$

$5.ln|x|-\frac{34}{3\sqrt{11}}.arctg\left(\frac{x+\frac{1}{6}}{\sqrt{\frac{11}{36}}}\right)-\frac{21}{4}ln\left|\left(x+\frac{1}{6}\right)^2+\frac{11}{46}\right|\right)$

Confira os calculos!

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