Question

integral

ache a integral de

integral (2x+1) / (3x+2) dx

Answer 1

Encontrar a integral indefinida:

$\large\begin{array}{l} \mathsf{I=\displaystyle\int\frac{2x+1}{3x+2}\,dx}\\\\ \mathsf{=\displaystyle\int 2\cdot \frac{x+\frac{1}{2}}{3x+2}\,dx}\\\\ \mathsf{=\displaystyle\int 2\cdot \dfrac{\,3\,}{3}\cdot \frac{x+\frac{1}{2}}{3x+2}\,dx}\\\\ \mathsf{=\displaystyle\int \frac{\,2\,}{3}\cdot \frac{3(x+\frac{1}{2})}{3x+2}\,dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int \frac{3x+\frac{3}{2}}{3x+2}\,dx} \end{array}$

$\large\begin{array}{l} \mathsf{=\displaystyle\frac{\,2\,}{3}\int \frac{3x+2-2+\frac{3}{2}}{3x+2}\,dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int\bigg(\frac{3x+2}{3x+2}+\frac{-2+\frac{3}{2}}{3x+2}\bigg)dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!\bigg(1+\frac{\frac{-4}{2}+\frac{3}{2}}{3x+2}\bigg)dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!\bigg(1+\frac{\frac{-1}{2}}{3x+2}\bigg)dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!\bigg(1-\frac{\,1\,}{2}\cdot \frac{1}{3x+2}\bigg)dx} \end{array}$

$\large\begin{array}{l} \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!\bigg(1-\frac{\,1\,}{2}\cdot \frac{\,3\,}{3}\cdot \frac{1}{3x+2}\bigg)dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!\bigg(1-\frac{\,1\,}{6}\cdot \frac{3}{3x+2}\bigg)dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!1\,dx+\frac{\,2\,}{3}\cdot \left(-\,\frac{\,1\,}{6}\right)\int\!\frac{3}{3x+2}\,dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!1\,dx-\frac{\,2\,}{18}\int\!\frac{1}{3x+2}\cdot 3\,dx} \end{array}$

$\large\begin{array}{l} \mathsf{=\displaystyle\frac{\,2\,}{3}\int\!1\,dx-\frac{\,1\,}{9}\int\!\frac{1}{3x+2}\cdot 3\,dx}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\,x-\frac{\,1\,}{9}\int\frac{\,1\,}{u}\,du\qquad(u=3x+2)}\\\\ \mathsf{=\displaystyle\frac{\,2\,}{3}\,x-\frac{\,1\,}{9}\,\ell n|u|+C}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\frac{2x+1}{3x+2}\,dx=\frac{\,2\,}{3}\,x-\frac{\,1\,}{9}\,\ell n\,|3x+2|+C} \end{array}}\qquad\checkmark \end{array}$


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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: integral indefinida função racional frações parciais cálculo integral

Answer 2

$\int \frac{2x + 1}{3x + 2} \: dx \\ = \int \frac{2x}{3x + 2} \: dx +\int \frac{1}{3x + 2}dx$

$\int \frac{2x}{3x + 2} \: dx \\ t = 3x + 2 \\ x = \frac{t - 2}{3} \\ dt = \frac{1}{3}dx$

$\int \frac{2x}{3x + 2}dx = \int \frac{2. \frac{t - 2}{3} }{t} . \frac{1}{3}dt \\ = \frac{2}{9} \int \frac{t - 2}{t} dt$

$= \frac{2}{9} \int \frac{t}{t}dt - \frac{4}{9}\int \frac{dt}{t} \\ = \frac{2}{9}t - \frac{4}{9} ln |t| + c$

$\frac{2}{9}(3x + 2) - \frac{4}{9} ln |3x + 2| + c$

$\int \frac{dx}{3x + 2} \\ h = 3x + 2 \\ dh = 3dx \\ dx = \frac{1}{3}dh$

$\int \frac{dx}{3x + 2} = \int \frac{ \frac{1}{3}dh}{h} = \frac{1}{3} \int \frac{dh}{h} \\ = \frac{1}{3} ln |h| + c = \frac{1}{3} ln |3x + 2| + c$

$\frac{2}{9}(3x + 2) - \frac{4}{9} ln |3x + 2| \\ + \frac{1}{3} ln |3x + 2|$

$= \frac{2}{9}(3x + 2) - \frac{1}{9}ln |3x + 2| + c$

$\int \frac{2x + 1}{3x + 2} \\ = \frac{1}{9}(3x + 2) - \frac{1}{9} ln |3x + 2| + c$