[tex]\int \, \frac{1}{2x+1} \, dx\\\\= \frac{1}{2} + \ln(|2x+1|) + C[/tex]
Em que C pertence ao conjunto dos Reais.
✅ Após realizar os cálculos, concluímos que a integral indefinida - antiderivada ou primitiva - procurada é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf \int \frac{1}{2x + 1}\,dx= \frac{1}{2}\cdot \ln|2x + 1| + c\:\:\:}}\end{gathered}$}[/tex]
Seja a integral "I":
[tex]\Large\displaystyle\text{$\begin{gathered}\tt \int \frac{1}{2x + 1}\,dx\end{gathered}$}[/tex]
Nesta situação, a função que devemos calcular sua integral indefinida é:
[tex]\Large\displaystyle\text{$\begin{gathered}\tt f(x) = \frac{1}{2x + 1}\end{gathered}$}[/tex]
Para resolver esta integral indefinida, devemos utilizar o método de substituição. Para isso, devemos:
[tex]\Large\displaystyle\text{$\begin{gathered}\tt u = 2x + 1\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\tt \frac{du}{dx} = 2\cdot1\cdot x^{1 - 1} + 0 = 2\cdot x^{0} = 2\cdot 1 = 2\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\tt \frac{du}{dx} = 2\Longrightarrow du = 2\,dx\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\tt \frac{du}{2} = dx\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\tt \int \frac{1}{u}\cdot\frac{du}{2} = \frac{1}{2}\int \frac{1}{u}\,du\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\tt = \frac{1}{2}\cdot\ln|u| + c\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\tt = \frac{1}{2}\cdot \ln|2x + 1| + c\end{gathered}$}[/tex]
✅ Portanto, a integral indefinida procurada é:
[tex]\Large\displaystyle\text{$\begin{gathered}\tt \int \frac{1}{2x + 1}\,dx= \frac{1}{2}\cdot \ln|2x + 1| + c\end{gathered}$}[/tex]
Saiba mais:
Solução gráfica (Figura):
