Intgerais

por **Manoella** » Qui Dez 16, 2010 09:44

Olá! Preciso que alguém mim explica sobre essas integrais ai;

a) $F\frac{1}{{(3x-5)}^{8}}dx$

b) $\int_{1}^{-1}\frac{{x}^{2}}{\sqrt[]{{x}^{3}+9}}dx$

por **Moura** » Qui Dez 16, 2010 10:54

u=3x-5 => du/dx=3 => dx=du/3

$\int_{}^{}\frac{1}{(3x-5)^8}dx$

$\int_{}^{}\frac{1}{u^8}dx=$ $\int_{}^{}{u}^{-8}dx=$ $\int_{}^{}{u}^{-8}\frac{du}{3}=$

$\frac{1}{3}\int_{}^{}{u}^{-8}du=$ $\frac{1}{3}\frac{{u}^{-7}}{7}=$ $\frac{{u}^{-7}}{21}=$ $\frac{(3x-5)^{-7}}{21}=$

$\frac{-1}{21(3x-5)^{7}}$ :y:

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u=x^3+9 => du/dx=3x^2 => dx=du/3x^2

$\int_{1}^{-1}\frac{x^2}{\sqrt[]{x^3+9}}dx =$ $\int_{1}^{-1}\frac{x^2}{(x^3+9)^\frac{1}{2}}dx =$ $\int_{1}^{-1}\frac{x^2}{u^\frac{1}{2}}dx =$

$\int_{1}^{-1}x^2u^\frac{-1}{2}dx =$ $\int_{1}^{-1}x^2u^\frac{-1}{2}\frac{du}{3x^2} =$ $\frac{1}{3}\int_{1}^{-1}u^\frac{-1}{2}du =$

$\frac{1}{3}\frac{u^\frac{1}{2}}{\frac{1}{2}}]_{1}^{-1} =$ $\frac{2}{3}u^\frac{1}{2}]_{1}^{-1} =$ $\frac{2(x^3+9)^\frac{1}{2}}{3}]_{1}^{-1} =$

$\frac{2((-1)^3+9)^\frac{1}{2}}{3}-(\frac{2((1)^3+9)^\frac{^1}{2}}{3}) =$ $\frac{4*\sqrt[]{2}}{3}-\frac{2*\sqrt[]{10}}{3}$ :y:

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