Como eu resolvo a integral de 1 até 3 de (2x/(x²+9)) dx;
Anexos:
![](https://pt-static.z-dn.net/files/d7e/92979c11eb4e3efbcb5488007801a4d7.jpg)
Soluções para a tarefa
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Basta usar a substituição. Muito simples ,veja:
![\int\limits^3_1 { \frac{2x}{x^2+9} } \, dx \ \ \ \ \ \ \ \ \ u=x^2+9 \ \ \ du/dx=2x \ \ \ \ \ \ dx= \frac{du}{2x} \\ \\ \int\limits^3_1 { \frac{2x}{u } \, \frac{du}{2x} \\ \\ \int\limits^3_1 { \frac{2x}{x^2+9} } \, dx \ \ \ \ \ \ \ \ \ u=x^2+9 \ \ \ du/dx=2x \ \ \ \ \ \ dx= \frac{du}{2x} \\ \\ \int\limits^3_1 { \frac{2x}{u } \, \frac{du}{2x} \\ \\](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E3_1+%7B+%5Cfrac%7B2x%7D%7Bx%5E2%2B9%7D+%7D+%5C%2C+dx++%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+u%3Dx%5E2%2B9+%5C+%5C+%5C+du%2Fdx%3D2x+%5C+%5C+%5C+%5C+%5C+%5C+dx%3D+%5Cfrac%7Bdu%7D%7B2x%7D++%5C%5C++%5C%5C++%5Cint%5Climits%5E3_1+%7B+%5Cfrac%7B2x%7D%7Bu+%7D+%5C%2C++%5Cfrac%7Bdu%7D%7B2x%7D++%5C%5C++%5C%5C+)
![\int\limits^3_1 { \frac{du}{u } \\ \\ \int\limits^3_1 { \frac{du}{u } \\ \\](https://tex.z-dn.net/?f=%5Cint%5Climits%5E3_1+%7B+%5Cfrac%7Bdu%7D%7Bu+%7D++%5C%5C++%5C%5C++)
![ln(u) \ |_1^3 \\ \\ ln(x^2+9) \ |_1^3 \\ \\ ln(3^2+9)-ln(1^2+9) \\ \\ ln(18)-ln(10) \\ \\ ln( \frac{18}{9} ) \\ \\ ln( \frac{9}{5}) \\ \\ =0,58\ (aprox) ln(u) \ |_1^3 \\ \\ ln(x^2+9) \ |_1^3 \\ \\ ln(3^2+9)-ln(1^2+9) \\ \\ ln(18)-ln(10) \\ \\ ln( \frac{18}{9} ) \\ \\ ln( \frac{9}{5}) \\ \\ =0,58\ (aprox)](https://tex.z-dn.net/?f=ln%28u%29+%5C+%7C_1%5E3+%5C%5C++%5C%5C+ln%28x%5E2%2B9%29+%5C+%7C_1%5E3+%5C%5C++%5C%5C+ln%283%5E2%2B9%29-ln%281%5E2%2B9%29+%5C%5C++%5C%5C+ln%2818%29-ln%2810%29+%5C%5C++%5C%5C+ln%28+%5Cfrac%7B18%7D%7B9%7D+%29+%5C%5C++%5C%5C+ln%28+%5Cfrac%7B9%7D%7B5%7D%29+%5C%5C++%5C%5C+%3D0%2C58%5C++%28aprox%29)
Comenta aí depois .
Comenta aí depois .
MatheusToniolli:
ok, só queria msm ver o método. Vlw
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