Question

Calcule a integral indefinida

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

sendo λ, k constantes reais não-nulas.

Answer 1

Olá Lucas! Vamos apenas aplicar algumas substituições.

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

Daí:

$\displaystyle \frac{1}{\lambda} \cdot \int \frac{\sqrt{v}}{e^u} \, \, dv \\ \\ \\ \frac{1}{\lambda} \cdot \int \frac{\sqrt{v}}{v-k} \, \, dv \\ \\ \\ w= \sqrt{v} \, \, \hookrightarrow \, \, v=w^2 \\ \\ dw = \frac{1}{2\sqrt{v}} \, \, dv \\ \\ dv = 2 \sqrt{v} \, \, dw$

Temos:

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

Continuando:

$\displaystyle -\frac{2}{\lambda} \cdot \bigg( \int \frac{k}{k-w^2} \, \, dw - \int 1 \, \, dw \bigg) \\ \\ \\ -\frac{2}{\lambda} \cdot \bigg( \frac{k}{k} \cdot \int \frac{1}{1-\displaystyle \frac{w^2}{k}} \, \, dw - w \bigg) \\ \\ \\ -\frac{2}{\lambda} \cdot \bigg( \int \frac{1}{1-\displaystyle \bigg( \frac{1}{\sqrt{k}}w \bigg)^2 } \, \, dw - w \bigg) \\ \\ \\ x = \frac{1}{\sqrt{k}}w \\ \\ \\ dx = \frac{1}{\sqrt{k}} \, \, dw \\ \\ \\ dw = \sqrt{k} \, \, dx$

Portanto:

$\displaystyle -\frac{2}{\lambda} \cdot \bigg( \sqrt{k} \cdot \int \frac{1}{1-x^2} \, \, dx - w \bigg)$

A seguinte integral é muito simples de fazer, no entanto mostrarei logo a resposta direta dela, mas os passos para fazer é através de frações parciais, comparando os numeradores e solucionando o sistema linear gerado pelas constantes A e B.

$\displaystyle \int \frac{1}{1-x^2} \, \, dx = \frac{1}{2} \ln |x+1| - \frac{1}{2} \ln |x-1|+c$

Daí ainda temos:

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

Só teremos que substituir os valores e simplificar:

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

==========

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

===========

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

Answer 2

É dada a integral:

$\displaystyle I=\int \sqrt{e^{\lambda x}+k}\,dx$

A princípio, eu resolveria da mesma forma que o colega que respondeu anteriormente. Porém, como essa solução já estava aqui, pensei em outro modo. Vamos fazer apenas uma substituição:

$\sqrt{e^{\lambda x}+k}=t+\sqrt{k}\Longrightarrow t=\sqrt{e^{\lambda x}+k}-\sqrt{k}$

Elevando os dois lados da primeira igualdade ao quadrado:

$e^{\lambda x}+k=t^2+2\sqrt k t+k\\\\ e^{\lambda x}=t^2+2\sqrt k t\\\\ \lambda e^{\lambda x}\,dx=(2t+2\sqrt k)\,dt\\\\ \lambda (t^2+2\sqrt k t)\,dx=2(t+\sqrt k)\,dt\\\\ dx=\dfrac{2}{\lambda}\cdot\dfrac{t+\sqrt k}{t^2+2\sqrt k t}\,dt$

Aplicando na integral:

$\displaystyle I=\int \sqrt{e^{\lambda x}+k}\,dx\\\\ I=\int (t+\sqrt{k})\cdot\dfrac{2}{\lambda}\cdot\dfrac{t+\sqrt k}{t^2+2\sqrt k t}\,dt\\\\ I=\dfrac{2}{\lambda}\int \dfrac{(t+\sqrt{k})^2}{t^2+2\sqrt k t}\,dt\\\\ I=\dfrac{2}{\lambda}\int \dfrac{t^2+2\sqrt k t+ k}{t^2+2\sqrt k t}\,dt\\\\ I=\dfrac{2}{\lambda}\int \left(\dfrac{t^2+2\sqrt k t}{t^2+2\sqrt k t}+\dfrac{k}{t^2+2\sqrt k t}\right)\,dt\\\\ I=\dfrac{2}{\lambda}\int dt+\dfrac{2k}{\lambda}\int\dfrac{1}{t^2+2\sqrt k t}\right)\,dt$
$\displaystyle I=\dfrac{2}{\lambda}t+\dfrac{2k}{\lambda}\int\underbrace{\dfrac{1}{t(t+2\sqrt k)}\right)}_{E}\,dt\\\\ E=\dfrac{1}{t(t+2\sqrt k)}=\dfrac{A}{t}+\dfrac{B}{t+2\sqrt k}=\dfrac{(A+B)t+2\sqrt k A}{t(t+2\sqrt k)}\\\\\\ \Longrightarrow \begin{cases}2\sqrt{k}A=1\Longrightarrow \underline{\overline{A=1/2\sqrt k}}\\A+B=0\Longrightarrow \underline{\overline{B=-1/2\sqrt k}}\end{cases}\\\\\\ I=\dfrac{2}{\lambda}t+\dfrac{2k}{\lambda}\int\left(\dfrac{1/2\sqrt k}{t}+\dfrac{-1/2\sqrt k}{t+2\sqrt k}\right)dt$
$\displaystyle I=\dfrac{2}{\lambda}t+\dfrac{\sqrt k}{\lambda}\int\dfrac{1}{t}dt-\dfrac{\sqrt k}{\lambda}\int\dfrac{1}{t+2\sqrt k}dt\\\\ I=\dfrac{2}{\lambda}t+\dfrac{\sqrt k}{\lambda}\ln|t|-\dfrac{\sqrt k}{\lambda}\ln|t+2\sqrt k|+C_0$

Agora, basta retornarmos à variável x:

$I=\dfrac{2}{\lambda}(\sqrt{e^{\lambda x}+k}-\sqrt{k})+\dfrac{\sqrt k}{\lambda}\ln|\sqrt{e^{\lambda x}+k}-\sqrt{k}|-\\-\dfrac{\sqrt k}{\lambda}\ln|(\sqrt{e^{\lambda x}+k}-\sqrt{k})+2\sqrt k|+C_0\\\\ I=\dfrac{2}{\lambda}\sqrt{e^{\lambda x}+k}-\underbrace{\dfrac{2}{\lambda}\sqrt{k}}_{C_1}+\dfrac{\sqrt k}{\lambda}\ln|\sqrt{e^{\lambda x}+k}-\sqrt{k}|-\\-\dfrac{\sqrt k}{\lambda}\ln|\sqrt{e^{\lambda x}+k}+\sqrt k|+C_0$

Considerando $C=C_0+C_1$, chegamos à resposta final:

$\boxed{I\!=\!\dfrac{2}{\lambda}\sqrt{e^{\lambda x}+k}+\!\dfrac{\sqrt k}{\lambda}\ln|\sqrt{e^{\lambda x}\!+\!k}-\!\sqrt{k}|-\dfrac{\sqrt k}{\lambda}\ln|\sqrt{e^{\lambda x}\!+\!k}+\!\sqrt k|\!+\!C}$