Question

Calcule a seguinte integral:
$\int\limits {\sqrt{1+e^{2x} } } \, dx$

Answer 1

• O resuldado da integral é igual a:

$\Large\displaystyle\text{ \begin{gathered}\sf \frac{1}{2}\left[\ln |\sqrt{1+e^{2x}} -1| - \ln|\sqrt{1+e^{2x}} +1|\right] + \sqrt{1+e^{2x}}+\mathbb{C}\end{gathered} }$

Desejamos calcular a seguinte integral:

$\Large\displaystyle\text{ \begin{gathered}\sf \int \sqrt{1+e^{2x}}\ dx \end{gathered} }$

Vamos fazer uma substituição, chamando então:

$\Large\displaystyle\begin{gathered}\sf u=e^{2x} \end{gathered}$

$\Large\displaystyle\begin{gathered}\sf du=2e^{2x}dx \end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \frac{du}{2u}=dx \end{gathered}$

Substituindo:

$\Large\displaystyle\text{ \begin{gathered}\sf \int \sqrt{1+e^{2x}}\ dx =\sf \int\frac{ \sqrt{1+u}}{2u}\ du \end{gathered} }$

E pela lineariedade, temos que:

$\Large\displaystyle\text{ \begin{gathered}\sf \int\frac{ \sqrt{1+u}}{2u}\ du =\frac{1}{2}\cdot\int\frac{ \sqrt{1+u}}{u}\ du \end{gathered} }$

Façamos então outra substituição, chamando então:

$\Large\displaystyle\begin{gathered}\sf c=\sqrt{1+u} \ \ \Rightarrow \ \ u=c^2-1\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\sf dc=\frac{1}{2\sqrt{1+u}} \ du\end{gathered} }$

$\Large\displaystyle\begin{gathered}\sf 2c\ dc= du\end{gathered}$

Ficando então:

$\Large\displaystyle\text{ \begin{gathered}\sf \frac{1}{2}\cdot\int\frac{ \sqrt{1+u}}{u}\ du =\frac{1}{2}\cdot \int \frac{c}{c^2-1}\cdot 2c\ dc \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\sf \frac{1}{2}\cdot\int\frac{ \sqrt{1+u}}{u}\ du =\frac{1}{2}\cdot \int \frac{2c^2}{c^2-1} \ dc \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\sf \frac{1}{2}\cdot\int\frac{ \sqrt{1+u}}{u}\ du =\int \frac{c^2}{c^2-1} \ dc \end{gathered} }$

Bom, agora devemos fazer uma divisão de polinômios, após isso, temos que:

$\Large\displaystyle\begin{gathered}\sf \int \frac{c^2}{c^2-1} \ dc = \int \frac{1}{c^2-1}+1\ dc\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \int \frac{c^2}{c^2-1} \ dc = \int \frac{1}{c^2-1}\ dc+\int dc\end{gathered}$

Sabemos que a integral de dc é igual a c, mas a outra teremos que abrir em frações parciais, logo:

$\Large\displaystyle\begin{gathered}\sf \frac{1}{(c-1)(c+1)}=\frac{A}{(c-1)} +\frac{B}{(c+1)} \end{gathered}$

$\Large\displaystyle\begin{gathered}\sf A(c+1)+B(c-1)=1\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf Ac+A+Bc-B=1\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf Ac + A + Bc - B-1=0\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf c(A+B)+(A-B-1)=0 \end{gathered}$

E com isso temos o seguinte sistema:

$\Large\displaystyle\begin{gathered}\begin{cases} \sf A+B=0\ \ \ \green{(I)}.\\ \sf A-B=1\ \ \ \green{(II)}.\end{cases}\end{gathered}$

Resolvendo:

$\Large\displaystyle\begin{gathered}\sf A=-B\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf -B-B=1\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\therefore \underline{\boxed{\sf B=-\frac{1}{2}}}\ \ \wedge\ \ \underline{\boxed{\sf A=\frac{1}{2}}}\end{gathered} }$

Substituindo:

$\Large\displaystyle\begin{gathered}\sf \int \frac{c^2}{c^2-1} \ dc = \left[\int \frac{1}{2(c-1)} \ dc - \int \frac{1}{2(c+1)}\right] + c\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \int \frac{c^2}{c^2-1} \ dc = \frac{1}{2}\left[\int \frac{1}{(c-1)} \ dc - \int \frac{1}{(c+1)}\ dc\right] + c\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \int \frac{c^2}{c^2-1} \ dc = \frac{1}{2}\left[\ln |c-1| - \ln|c+1|\right] + c\end{gathered}$

Agora é só ir voltando com as substituições que fizemos, logo:

$\Large\displaystyle\begin{gathered}\sf = \frac{1}{2}\left[\ln |\sqrt{1+u} -1| - \ln|\sqrt{1+u} +1|\right] + \sqrt{1+u} \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\sf = \frac{1}{2}\left[\ln |\sqrt{1+e^{2x}} -1| - \ln|\sqrt{1+e^{2x}} +1|\right] + \sqrt{1+e^{2x}}\end{gathered} }$

E essa é a resposta da nossa integral, basta por fim adicionar uma constante.

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