Question

Calcule a integral das funções:
PotF(x)=2 ln⁡(x+1)+4
PotR(x)=1/e^x +3

Answer 1

Os valores das integrais são calculados a seguir.

Explicação passo-a-passo:

$F(x)=\int{(2\,.\,ln\;(x+1)+4)\;dx}\\\\F(x)=\int{2\,.\,ln\;(x+1)\;dx}+\int{4\;dx}\\\\F(x)=2\int{ln\;(x+1)\;dx}+4\int{1\;dx}\\\\F(x)=2\int{ln\;(x+1)\;dx}+4x+C\\\\u=ln\;(x+1)\quad\quad\quad dv=dx\\\\du=\frac{1}{x+1}\;dx\quad\quad\quad v=x\\\\F(x)=2\left[u\,.\,v-\int{v\;du}\right]+4x+C\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-\int{x\,.\,\frac{1}{x+1}\;dx}\right]+4x+C\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-\int{\frac{x}{x+1}\;dx}\right]+4x+C$

$u=x+1\;\rightarrow\; \frac{du}{dx}=1\;\rightarrow\;dx=du\\\\u=x+1\;\rightarrow\;x=u-1\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-\int{\frac{u-1}{u}\;du}\right]+4x+C\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-\int{(1-\frac{1}{u})\;du}\right]+4x+C\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-(\int{1\;du}-\int{\frac{1}{u}\;du)}\right]+4x+C\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-(u-ln\;u)}\right]+4x+C\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-u+ln\;u}\right]+4x+C$

$F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-(x+1)+ln\;(x+1)}\right]+4x+C\\\\F(x)=2\,.\,\left[ln\;(x+1)\,.\,x-x-1+ln\;(x+1)}\right]+4x+C\\\\F(x)=2\,.\,\left[(x+1)\,.\,ln\;(x+1)-x-1}\right]+4x+C\\\\F(x)=2\,.\,(x+1)\,.\,ln\;(x+1)-2x-2+4x+C\\\\F(x)=2\,.\,(x+1)\,.\,ln\;(x+1)+2x-2+C\\\\F(x)=2\,.\,\left[(x+1)\,.\,ln\;(x+1)+x \right]+C$

$R(x)=\int {\frac{1}{e^x}+3}\;dx\\\\R(x)=\int {\frac{1}{e^x}}\;dx+\int{3}\;dx\\\\R(x)=\int {e^{-x}}}\;dx+3\,.\,\int\;dx\\\\R(x)=-e^{-x}+3x+C\\\\R(x)=3x-e^{-x}+C$