Question

Resolva as integrais

Answer 1

$\boxed{\boxed{\int\limits^1_0 { \frac{x}{x^2+1} } \, dx }}$

integrando por substituição
$u = x^2+1$

derivando em relaçao ao x
$\frac{du}{dx}= 2x \\\\du=2x.dx\\\\ \frac{du}{2x}=dx$

substituindo os valores
$\int\limits^1_0 { \frac{\not x}{u} } \, \frac{dx}{2\not x} =\\\\ = \int\limits^1_0 { \frac{1}{2u} } \, du \\\\= \frac{1}{2} \int\limits^1_0 { \frac{1}{u} } \, du$

integrando
$\int\limits^1_0 { \frac{1}{u} } \, du = |ln(u) |^1_0 = |ln(x^2+1)|^1_0= [ln(1^2+1) - ln(0^2+1)] = ln(2)$

temos
$\frac{1}{2}*ln(2) = \frac{ln(2)}{2}$
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$\int\limits^2_1 {x*e^{-x^2+1}} \, dx$

integrando por substituiçao
$u = -x^2+1\\\\ \frac{du}{dx}= -2x \to \frac{du}{-2x} =dx$

.
$\int\limits^2_1 {x*e^u} \, \frac{du}{-2x} \\\\ \int\limits^2_1 {e^u} \, \frac{du}{-2} \\\\ \frac{-1}{2} \int\limits^2_1 {e^u} \, du \\\\= \frac{-1}{2} [e^u]^2_1\\\\= \frac{-1}{2} [ e^{-(x)^2+1}]^1_0\\\\= \frac{-1}{2}[e^{-(2)^2+1} - e^{-1()^2+1}]\\\\= \frac{-1}{2}[e^{-3}-e^{0}}] = \frac{-e^{3}+1}{2}$
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a ultima não da pra enxergar direito os expoentes