Question

Calcule a integral indefinida:

$\displaystyle\int\frac{1}{\sqrt{e^{\lambda x}+k^2}}\,dx$

onde λ, k são constantes, λ ≠ 0, k > 0.

Answer 1

Olá Lucas! Vamos fazer algumas substituições e alterações algébricas. Relaxe e olhe:

$\displaystyle \mathsf{ \int \frac{1}{\sqrt{e^{\lambda x} + k^2}} \, dx } \\ \\ \\ \mathsf{u= \lambda x} \\ \\ \mathsf{du = \lambda \, dx} \\ \\ \\ \mathsf{\frac{1}{\lambda} \cdot \int \frac{1}{\sqrt{e^u + k^2}} \, du} \\ \\ \\ \mathsf{v=\sqrt{e^u + k^2} \, \, \, \hookrightarrow \, \, \, e^u = v^2-k^2} \\ \\ \mathsf{dv = \frac{e^u}{2 \sqrt{e^u+k^2}} \, du} \\ \\ \\ \mathsf{du = \frac{2\sqrt{e^u+k^2}}{e^u} \, dv}$

Continuando em outra linha:

$\displaystyle \mathsf{\frac{2}{\lambda} \cdot \int \frac{\sqrt{e^u+k^2}}{e^u \cdot v} \, dv} \\ \\ \\ \mathsf{\frac{2}{\lambda} \cdot \int \frac{\sqrt{v^2-k^2+k^2}}{(v^2-k^2) \cdot v} \, dv} \\ \\ \\ \mathsf{\frac{2}{\lambda} \cdot \int \frac{v}{(v^2-k^2) \cdot v} \, dv} \\ \\ \\ \mathsf{ \frac{2}{\lambda} \cdot \int \frac{1}{v^2-k^2} \, dv }$

Agora vamos puxar a constante para fora do integrando:

$\displaystyle \mathsf{ \frac{2}{\lambda} \cdot \frac{1}{k^2} \cdot \int \frac{1}{\displaystyle \frac{v^2}{k^2}-1} \, dv } \\ \\ \\ \mathsf{\frac{2}{\lambda k^2} \cdot \int \frac{1}{\displaystyle \bigg( \frac{v}{k} \bigg)^2 -1 } \, dv} \\ \\ \\ \mathsf{\frac{2}{\lambda k^2} \cdot \int \frac{1}{\displaystyle \bigg( \frac{1}{k}v \bigg)^2 -1 } \, dv} \\ \\ \\ \mathsf{w=\frac{1}{k}v} \\ \\ \\ \mathsf{dw = \frac{1}{k} \, dv} \\ \\ \\ \mathsf{dv=k \, dw}$

Daí temos:

$\displaystyle \mathsf{ \frac{2}{\lambda k^2} \cdot k \cdot \int \frac{1}{w^2-1} \, dw} \\ \\ \\ \mathsf{ \frac{2}{\lambda k} \cdot \int \frac{1}{w^2-1} \, dw}$

Todos nós já estamos familiarizados com essa integral. Mostrarei a resposta diretamente, mas a resolução dela se dá utilizando frações parciais. Para não perdermos o foco, temos:

$\displaystyle \mathsf{ \frac{2}{\lambda k} \cdot \bigg( \frac{1}{2} \ln |w-1| - \frac{1}{2} \ln |w+1| \bigg) + C} \\ \\ \\ \mathsf{\frac{1}{\lambda k} \ln |w-1| -\frac{1}{\lambda k} \ln |w+1| + C } \\ \\ \\ \mathsf{ \frac{1}{\lambda k} \ln \bigg|\frac{v}{k}-1 \bigg| -\frac{1}{\lambda k} \ln \bigg|\frac{v}{k}+1 \bigg| + C } \\ \\ \\ \mathsf{\frac{1}{\lambda k} \ln \bigg|\frac{\sqrt{e^u+k^2}}{k}-1 \bigg| -\frac{1}{\lambda k} \ln \bigg|\frac{\sqrt{e^u+k^2}}{k}+1 \bigg| + C}$

Continuando:

$\displaystyle \mathsf{\frac{1}{\lambda k} \ln \bigg|\frac{\sqrt{e^{\lambda x}+k^2}}{k}-1 \bigg| -\frac{1}{\lambda k} \ln \bigg|\frac{\sqrt{e^{\lambda x}+k^2}}{k}+1 \bigg| + C} \\ \\ \\ \mathsf{\frac{1}{\lambda k} \ln \bigg| \frac{\sqrt{e^{\lambda x}+k^2}-k}{k} \bigg| - \frac{1}{\lambda k} \ln \bigg| \frac{\sqrt{e^{\lambda x}+k^2} + k}{k} \bigg| + C}$

Colocando o termo em evidência, temos:

$\displaystyle \mathsf{\frac{1}{\lambda k} \cdot \bigg( \ln \bigg| \frac{\sqrt{e^{\lambda x}+k^2}-k}{k} \bigg| - \ln \bigg| \frac{\sqrt{e^{\lambda x}+k^2} + k}{k} \bigg| \bigg) + C}$

De acordo com a seguinte propriedade de logaritmos:

$\displaystyle \mathsf{ \ln \bigg| \, \frac{x}{y} \, \bigg| = \ln |x|- \ln |y| }$

Daí:

$\displaystyle \mathsf{ \frac{1}{\lambda k} \cdot \Bigg( \ln \bigg| \sqrt{e^{\lambda x} + k^2} - k \bigg| - ln |k| - \bigg( \ln \bigg| \sqrt{e^{\lambda x} + k^2} + k \bigg| - \ln |k| \bigg) \Bigg)+ C} \\ \\ \\ \mathsf{ \frac{1}{\lambda k} \cdot \bigg( \ln \bigg| \sqrt{e^{\lambda x} + k^2} - k \bigg| - \ln |k| - \ln \bigg| \sqrt{e^{\lambda x} + k^2} + k \bigg| + \ln |k| \bigg)+ C}$

Continuando o resultado final em outra linha, temos:

$\displaystyle \mathsf{ \frac{1}{\lambda k} \cdot \bigg( \ln \bigg| \sqrt{e^{\lambda x} + k^2} - k \bigg| - \ln \bigg| \sqrt{e^{\lambda x} + k^2} + k \bigg| \bigg)+ C} \\ \\ \\ \mathsf{ \frac{1}{\lambda k} \ln \bigg| \sqrt{e^{\lambda x} + k^2} - k \bigg| - \frac{1}{\lambda k} \ln \bigg| \sqrt{e^{\lambda x} + k^2} + k \bigg| + C} \\ \\ \\ \hookrightarrow \, \, \mathsf{\frac{\ln |\sqrt{e^{\lambda x} + k^2} -k | - \ln | \sqrt{e^{\lambda x} + k^2} +k | }{\lambda k} + C}$

Answer 2

Olá Lukyo.


Multiplique em cima e embaixo a função que queremos integrar por $\mathsf{\lambda e^{\lambda x}}$

$\mathsf{\displaystyle\int \dfrac{\lambda e^{\lambda x}}{\lambda e^{\lambda x}} \cdot\dfrac{1}{\sqrt{e^{\lambda x}+k^2}}~dx}$


Faça:


$\mathsf{u=\sqrt{e^{\lambda x}+k^2}}}\\\\\\\mathsf{u^2=e^{\lambda x}+k^2}\\\\\\\mathsf{2u~du=\lambda\cdot e^{\lambda x}~dx}$

Substituindo:

$\mathsf{\displaystyle\dfrac{1}{\lambda}\cdot\int \dfrac{2u}{(u^2-k^2)\cdot u}~du= \dfrac{2}{\lambda}\cdot\int\dfrac{1}{u^2-k^2}~du}$

A partir daqui iremos decompor a nossa integral em frações parciais.

Note que:

$\mathsf{\dfrac{1}{u^2-k^2}=\dfrac{1}{(u-k)(u+k)}}$

A ideia é considerar a existência de duas constantes A e B tais que:

$\mathsf{\dfrac{1}{(u-k)(u+k)}=\dfrac{A}{u-k}+\dfrac{B}{u+k}}$

Desenvolvendo

$\mathsf{\dfrac{1}{(u-k)(u+k)}=\dfrac{A(u+k)+B(u-k)}{(u-k)(u+k)}=\dfrac{u(A+B)+k(A-B)}{(u-k)(u+k)}}$

Basta resolver o seguinte sistema


$\mathsf{\begin{cases}\mathsf{A+B=0}\\\\\mathsf{A-B=\dfrac{1}{k}}\end{cases}~~\Rightarrow~~A=\dfrac{1}{2k}~~~e~~~B=-\dfrac{1}{2k}}$


Portanto, temos:


$\mathsf{\dfrac{1}{u^2-k^2}=\dfrac{1}{2k(u-k)}}-\dfrac{1}{2k(u+k)}}$


Substituindo.


$\mathsf{\displaystyle\dfrac{2}{\lambda}\cdot\int \dfrac{1}{u^2-k^2}~du=\dfrac{2}{2k\lambda}\cdot\int \dfrac{1}{u-k}~du-\dfrac{2}{2k\lambda}\cdot\int\dfrac{1}{u+k}~du}\\\\\\\\\mathsf{\displaystyle\dfrac{2}{\lambda}\cdot\int \dfrac{1}{u^2-k^2}~du=\dfrac{1}{k\lambda}\cdot(\ell n|u-k|-\ell n|u+k|)+C}$


Substituindo u.


$\mathsf{\displaystyle\int \dfrac{1}{\sqrt{e^{\lambda x}+k^2}}~dx=\dfrac{1}{k\lambda}\cdot(\ell n|\sqrt{e^{\lambda x}+k^2}-k|-\ell n|\sqrt{e^{\lambda x}+k^2}+k|)+C}$


Dúvidas? comente.