Question

calcule a integral indefinida

Answer 1

Seja

$u=x^2+1. \\ \\ \dfrac{du}{dx}=2x$

Então,

$\int \dfrac{2x}{x^2+1} dx = \int \dfrac{du}{dx} \cdot \dfrac{1}{u} dx = \int \dfrac{du}{u}=log(u)+C=log(x^2 + 1) + C$

Answer 2

✅ Após resolver os cálculos, concluímos que a primitiva ou antiderivada ou integral indefinida da referida função racional é:

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf F(x) = \int \left(\frac{2x}{1 + x^{2}}\right)\,dx = \ln(x^{2} + 1) + c\:\:\:}}\end{gathered} }$

Seja a integral indefinida:

                   $\LARGE\displaystyle\text{ \begin{gathered}\tt \int\frac{2x}{1 + x^{2}}\,dx\end{gathered} }$

Observe que a integral indefinida é gerada a partir da seguinte função racional:

                 $\LARGE\displaystyle\text{ \begin{gathered}\tt f(x) = \frac{2x}{1 + x^{2}}\end{gathered} }$

Para resolver esta integral indefinida devemos fazer:

   $\LARGE\displaystyle\begin{gathered}\tt u = 1 + x^{2} = x^{2} + 1\end{gathered}$

   $\LARGE\displaystyle\begin{gathered}\tt \frac{du}{dx} = 2x\end{gathered}$

Desta forma temos a seguinte primitiva:

    $\LARGE \text { \begin{aligned}\tt F(x) & = \tt \int f(x)\,dx\\\tt & = \tt \int \left(\frac{2x}{1 + x^{2}}\right)\,dx\\\tt & = \tt \int \left(\frac{2x}{x^{2} + 1}\right)\,dx\\\tt & = \tt \int \frac{du}{dx}\cdot\frac{1}{u}\,dx\\\tt & = \tt \int \frac{du}{u}\\\tt & = \tt \ln(u) + c\\\tt & = \tt \ln(x^{2} + 1) + c\end{aligned} }$

✅ Portanto, a integral indefinida é:

    $\LARGE\displaystyle\text{ \begin{gathered}\tt F(x) = \int \left(\frac{2x}{1 + x^{2}}\right)\,dx = \ln(x^{2} + 1) + c\end{gathered} }$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$