Question

∫ $\frac{e ^{x} x}{1-e ^{2x} }$

Answer 1

Cálculo de integral indefinida:

$\large\begin{array}{l} \mathsf{\displaystyle\int\!\frac{e^x}{1-e^{2x}}\,dx}\\\\ =\mathsf{\displaystyle\int\!\frac{e^x}{1-(e^x)^2}\,dx}\\\\ =\mathsf{\displaystyle\int\!\frac{1}{(1-e^x)(1+e^x)}\cdot e^x\,dx\qquad\quad{(i)}}\\\\ \end{array}$


$\large\begin{array}{l} \textsf{Fa\c{c}a a seguinte substitui\c{c}\~ao:}\\\\ \mathsf{e^x=u~~\Rightarrow~~e^x\,dx=du}\\\\\\ \textsf{e a integral (i) fica} \end{array}$

$\large\begin{array}{l} \mathsf{\displaystyle\int\!\frac{1}{(1-u)(1+u)}\,du\qquad\quad(ii)} \end{array}$

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$\large\begin{array}{l} \textsf{Decompondo a fun\c{c}\~ao integrando em fra\c{c}\~oes parciais:}\\\\ \mathsf{g(u)=\dfrac{1}{(1-u)(1+u)}=\dfrac{A}{1-u}+\dfrac{B}{1+u}}\\\\ \mathsf{\dfrac{1}{(1-u)(1+u)}=\dfrac{A(1+u)+B(1-u)}{(1-u)(1+u)}}\\\\ \mathsf{\dfrac{1}{(1-u)(1+u)}=\dfrac{A+Au+B-Bu}{(1-u)(1+u)}}\\\\ \mathsf{\dfrac{1}{(1-u)(1+u)}=\dfrac{(A-B)u+A+B}{(1-u)(1+u)}} \end{array}$


$\large\begin{array}{l} \textsf{Por identidade polinomial dos numeradores, devemos ter}\\\\ \left\{\!\begin{array}{l} \mathsf{A-B=0}\\ \mathsf{A+B=1} \end{array}\right.\\\\\\ \textsf{Resolvendo o sistema acima pelo m\'etodo da adi\c{c}\~ao:}\\\\ \mathsf{2A=1}\\\\ \mathsf{A=\dfrac{1}{2}} \end{array}$


$\large\begin{array}{l} \mathsf{A-B=0}\\\\ \mathsf{B=A}\\\\ \mathsf{B=\dfrac{1}{2}}\\\\\\ \textsf{Ent\~ao a fun\c{c}\~ao a ser integrada \'e}\\\\ \mathsf{g(u)=\dfrac{\frac{1}{2}}{1-u}+\dfrac{\frac{1}{2}}{1+u}}\\\\ \mathsf{g(u)=\dfrac{-\,\frac{1}{2}}{u-1}+\dfrac{\frac{1}{2}}{u+1}} \end{array}$


$\large\begin{array}{l} \textsf{Ent\~ao, a integral (ii) fica}\\\\ =\mathsf{\displaystyle\int\!\bigg(\frac{-\,\frac{1}{2}}{u-1}+\frac{\frac{1}{2}}{u+1} \bigg)\,du}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\int\!\frac{1}{u-1}\,du+\frac{1}{2}\int\!\frac{1}{u+1}\,du}\\\\ =\mathsf{-\,\dfrac{1}{2}\,\ell n|u-1|+\dfrac{1}{2}\,\ell n|u+1|+C}\\\\ \boxed{\begin{array}{c}\mathsf{\displaystyle\int\!\frac{e^x}{1-e^{2x}}\,dx}=\mathsf{\displaystyle-\,\frac{1}{2}\,\ell n|e^x-1|+\frac{1}{2}\,\ell n(e^x+1)+C}\end{array}}\qquad\quad\checkmark \end{array}$


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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: integral indefinida exponencial substituição função racional frações parciais cálculo integral