Exercício:(Nível de dificuldade: baixo, mas trabalhoso...)
Determine uma expressão para
\[ \int {\frac{{2x^2 + 1}}{{x\left( {x^2 + 2x + 7} \right)^2}}dx} \]
\[ \frac{1}{{49}}\ln \left| x \right| - \frac{1}{{98}}\ln \left( x^2 + 2x + 7 \right) - \frac{{13\sqrt 6 }}{{392}}\arctan \left( {\left( {x + 1} \right)\frac{{\sqrt 6 }}{6}} \right) - \frac{{5x + 31}}{{28\left( x^2 + 2x + 7 \right)}} + c; c\in\R \]
Determine uma expressão para
\[ \int {\frac{{2x^2 + 1}}{{x\left( {x^2 + 2x + 7} \right)^2}}dx} \]
\[ \frac{1}{{49}}\ln \left| x \right| - \frac{1}{{98}}\ln \left( x^2 + 2x + 7 \right) - \frac{{13\sqrt 6 }}{{392}}\arctan \left( {\left( {x + 1} \right)\frac{{\sqrt 6 }}{6}} \right) - \frac{{5x + 31}}{{28\left( x^2 + 2x + 7 \right)}} + c; c\in\R \]
$x^2 + 2x + 7$ não tem zeros reais pois $x^2 + 2x + 7=(x+1)^2+6$.
Assim, a fracção decompõe-se da seguinte forma: \[ \frac{{2x^2 + 1}}{{x\left( {x^2 + 2x + 7} \right)^2 }} = \frac{A}{x} + \frac{{Bx + C}}{{x^2 + 2x + 7}} + \frac{{Dx + E}}{{\left( {x^2 + 2x + 7} \right)^2 }} \] \[ \Leftrightarrow \frac{{2x^2 + 1}}{{x\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} = \frac{{A\left( {\left( {x + 1} \right)^2 + 6} \right)^2 + \left( {Bx + C} \right)x\left( {x^2 + 2x + 7} \right) + x\left( {Dx + E} \right)}}{{x\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} \] \[ \Leftrightarrow 2x^2 + 1 = \left( {A + B} \right)x^4 + \left( {4A + 2B + C} \right)x^3 + \left( {18A + 7B + 2C + D} \right)x^2 + \left( {28A + 7C + E} \right)x + 49A \] \[ \Leftrightarrow \left\{ {\begin{array}{ccl} 0 & = & {A + B} \\ 0 & = & {4A + 2B + C} \\ 2 & = & {18A + 7B + 2C + D} \\ 0 & = & {28A + 7C + E} \\ 1 & = & {49A} \end{array}} \right. \] \[ \Leftrightarrow \left\{ {\begin{array}{ccl} { - \displaystyle\frac{1}{{49}}} & = & B \\ { - \displaystyle\frac{2}{{49}}} & = & C \\ {\displaystyle\frac{{13}}{{7}}} & = & D \\ { - \displaystyle\frac{2}{{7}}} & = & E \\ {\displaystyle\frac{1}{{49}}} & = & A \end{array}} \right. \] Portanto \[ \frac{{2x^2 + 1}}{{x\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} = \frac{{\frac{1}{{49}}}}{x} + \frac{{ - \frac{1}{{49}}x - \frac{2}{{49}}}}{{\left( {x + 1} \right)^2 + 6}} + \frac{{\frac{{13}}{{7}}x - \frac{2}{{7}}}}{{\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} \] Atendendo a que: \begin{eqnarray*} {\int {\frac{{\frac{1}{{49}}}}{x}dx}}&{=}&{\frac{1}{{49}}\int {\frac{1}{x}dx} = \frac{1}{{49}}\ln \left| x \right| + C_1}\\ \end{eqnarray*} \begin{eqnarray*} {\int {\frac{\frac{13}{7}x - \frac{2}{7}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx}}&{=}&{\frac{1}{7}\left( {\int {\frac{{13x + 13 - 15}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } \right)}\\ {}&{=}&{- \frac{1}{{49}}\left( {\frac{1}{2}\int {\frac{{2x + 2}}{{x^2 + 2x + 7}}dx} + \int {\frac{1}{{\left( {x + 1} \right)^2 + 6}}dx} } \right)}\\ {}&{=}&{- \frac{1}{{49}}\left( {\frac{1}{2}\int {\frac{{2x + 2}}{{x^2 + 2x + 7}}dx} + \frac{\sqrt 6}{6} \int {\frac{{\frac{1}{{\sqrt 6 }}}}{{\left( {\frac{{x + 1}}{{\sqrt 6 }}} \right)^2 + 1}}dx} } \right)}\\ {}&{=}&{-\frac{1}{98}\ln {\left(x^2+7x+1\right)}-\frac{\sqrt 6}{294}\arctg{\frac{x+1}{\sqrt 6}}+C_2} \end{eqnarray*} E \begin{eqnarray*} {\int {\frac{{\frac{{13}}{7}x - \frac{2}{7}}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx}}&{ = }&{\frac{1}{7}\left( {\int {\frac{{13x + 13 - 15}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } \right)}\\ {}&{=}&{\frac{1}{2} \times \frac{{13}}{7}\int {\frac{{2x + 2}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx - \frac{{15}}{7}\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } }\\ {}&{=}&{ \frac{{13}}{{14}}\int {\left( {2x + 2} \right)\left( {x^2 + 2x + 7} \right)^{ - 2} dx - \frac{{15}}{7}\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } }\\ {}&{=}&{ - \frac{{13}}{{14}}\frac{1}{{\left( {x^2 + 2x + 7} \right)}} - \frac{{15}}{7}\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } \end{eqnarray*} e ainda \begin{eqnarray*} {\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx}}&{ = }&{\int{\frac{1}{((x+1)^2+6)^2}}dx}\\ \end{eqnarray*} Fazendo a susbstituição \[t=\arctg \left(\frac{x+1}{\sqrt{6}}\right)\] temos \begin{eqnarray*} {}&{ = }&{\int{\frac{\sqrt{6}\sec^2 t}{\left(6\sec^2t\right)^2}}dt}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\int{\cos^2 t}dt}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\int{\frac{1+\cos(2t)}{2}}dt}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{\sen(2t)}{4}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{2\sen(t)\cos(t)}{4}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{2\tg(t)\cos^2(t)}{4}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{\tg(t)}{2\left(1+\tg^2 t\right)}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{\arctg \left(\frac{x+1}{\sqrt{6}}\right)}{2}+\frac{\frac{x+1}{\sqrt{6}}}{2\left(1+\left(\frac{x+1}{\sqrt{6}}\right)^2\right)}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{72}\arctg \left( \frac{x+1}{\sqrt{6}} \right)+\frac{x+1}{12\left(x^2+2x+7\right)}+C_3 } \end{eqnarray*} Combinando estes resultados todos obtém-se a resposta apresentada.
As constantes $C_1,C_2$ e $C_3$ são números reais
Assim, a fracção decompõe-se da seguinte forma: \[ \frac{{2x^2 + 1}}{{x\left( {x^2 + 2x + 7} \right)^2 }} = \frac{A}{x} + \frac{{Bx + C}}{{x^2 + 2x + 7}} + \frac{{Dx + E}}{{\left( {x^2 + 2x + 7} \right)^2 }} \] \[ \Leftrightarrow \frac{{2x^2 + 1}}{{x\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} = \frac{{A\left( {\left( {x + 1} \right)^2 + 6} \right)^2 + \left( {Bx + C} \right)x\left( {x^2 + 2x + 7} \right) + x\left( {Dx + E} \right)}}{{x\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} \] \[ \Leftrightarrow 2x^2 + 1 = \left( {A + B} \right)x^4 + \left( {4A + 2B + C} \right)x^3 + \left( {18A + 7B + 2C + D} \right)x^2 + \left( {28A + 7C + E} \right)x + 49A \] \[ \Leftrightarrow \left\{ {\begin{array}{ccl} 0 & = & {A + B} \\ 0 & = & {4A + 2B + C} \\ 2 & = & {18A + 7B + 2C + D} \\ 0 & = & {28A + 7C + E} \\ 1 & = & {49A} \end{array}} \right. \] \[ \Leftrightarrow \left\{ {\begin{array}{ccl} { - \displaystyle\frac{1}{{49}}} & = & B \\ { - \displaystyle\frac{2}{{49}}} & = & C \\ {\displaystyle\frac{{13}}{{7}}} & = & D \\ { - \displaystyle\frac{2}{{7}}} & = & E \\ {\displaystyle\frac{1}{{49}}} & = & A \end{array}} \right. \] Portanto \[ \frac{{2x^2 + 1}}{{x\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} = \frac{{\frac{1}{{49}}}}{x} + \frac{{ - \frac{1}{{49}}x - \frac{2}{{49}}}}{{\left( {x + 1} \right)^2 + 6}} + \frac{{\frac{{13}}{{7}}x - \frac{2}{{7}}}}{{\left( {\left( {x + 1} \right)^2 + 6} \right)^2 }} \] Atendendo a que: \begin{eqnarray*} {\int {\frac{{\frac{1}{{49}}}}{x}dx}}&{=}&{\frac{1}{{49}}\int {\frac{1}{x}dx} = \frac{1}{{49}}\ln \left| x \right| + C_1}\\ \end{eqnarray*} \begin{eqnarray*} {\int {\frac{\frac{13}{7}x - \frac{2}{7}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx}}&{=}&{\frac{1}{7}\left( {\int {\frac{{13x + 13 - 15}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } \right)}\\ {}&{=}&{- \frac{1}{{49}}\left( {\frac{1}{2}\int {\frac{{2x + 2}}{{x^2 + 2x + 7}}dx} + \int {\frac{1}{{\left( {x + 1} \right)^2 + 6}}dx} } \right)}\\ {}&{=}&{- \frac{1}{{49}}\left( {\frac{1}{2}\int {\frac{{2x + 2}}{{x^2 + 2x + 7}}dx} + \frac{\sqrt 6}{6} \int {\frac{{\frac{1}{{\sqrt 6 }}}}{{\left( {\frac{{x + 1}}{{\sqrt 6 }}} \right)^2 + 1}}dx} } \right)}\\ {}&{=}&{-\frac{1}{98}\ln {\left(x^2+7x+1\right)}-\frac{\sqrt 6}{294}\arctg{\frac{x+1}{\sqrt 6}}+C_2} \end{eqnarray*} E \begin{eqnarray*} {\int {\frac{{\frac{{13}}{7}x - \frac{2}{7}}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx}}&{ = }&{\frac{1}{7}\left( {\int {\frac{{13x + 13 - 15}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } \right)}\\ {}&{=}&{\frac{1}{2} \times \frac{{13}}{7}\int {\frac{{2x + 2}}{{\left( {x^2 + 2x + 7} \right)^2 }}dx - \frac{{15}}{7}\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } }\\ {}&{=}&{ \frac{{13}}{{14}}\int {\left( {2x + 2} \right)\left( {x^2 + 2x + 7} \right)^{ - 2} dx - \frac{{15}}{7}\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } }\\ {}&{=}&{ - \frac{{13}}{{14}}\frac{1}{{\left( {x^2 + 2x + 7} \right)}} - \frac{{15}}{7}\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx} } \end{eqnarray*} e ainda \begin{eqnarray*} {\int {\frac{1}{{\left( {x^2 + 2x + 7} \right)^2 }}dx}}&{ = }&{\int{\frac{1}{((x+1)^2+6)^2}}dx}\\ \end{eqnarray*} Fazendo a susbstituição \[t=\arctg \left(\frac{x+1}{\sqrt{6}}\right)\] temos \begin{eqnarray*} {}&{ = }&{\int{\frac{\sqrt{6}\sec^2 t}{\left(6\sec^2t\right)^2}}dt}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\int{\cos^2 t}dt}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\int{\frac{1+\cos(2t)}{2}}dt}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{\sen(2t)}{4}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{2\sen(t)\cos(t)}{4}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{2\tg(t)\cos^2(t)}{4}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{t}{2}+\frac{\tg(t)}{2\left(1+\tg^2 t\right)}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{6^2}\left(\frac{\arctg \left(\frac{x+1}{\sqrt{6}}\right)}{2}+\frac{\frac{x+1}{\sqrt{6}}}{2\left(1+\left(\frac{x+1}{\sqrt{6}}\right)^2\right)}\right)+C_3}\\ {}&{=}&{\frac{\sqrt{6}}{72}\arctg \left( \frac{x+1}{\sqrt{6}} \right)+\frac{x+1}{12\left(x^2+2x+7\right)}+C_3 } \end{eqnarray*} Combinando estes resultados todos obtém-se a resposta apresentada.
As constantes $C_1,C_2$ e $C_3$ são números reais
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