Question

integrais - cálculo 1

Answer 1

$\int \left(9t^2+ \frac{1}{ \sqrt{t^3} } \right)dt\\\\ = \int \left(9t^2+ t^{- \frac{3}{2} }\right)dt\\\\= \frac{9t^{2+1}}{2+1} + \frac{t^{ \frac{-3}{2}+1 }}{ \frac{-3}{2}+1} +C\\\\=3t^3 + \frac{t^{- \frac{-1}{2}}}{ \frac{-1}{2}} +C\\\\=3t^3- \frac{2t^{ \frac{-1}{2}}}{1}\\\\= 3t^3 - \frac{2}{ \sqrt{t} } +C$

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$\int tg(x) dx=\int \frac{sen(x)}{cos(x)}dx$

substituiçao
U = cos(x)
dU = -sen(x) dx

$\int \frac{-1}{u}du = -ln(u) +C = -ln(cos(x))+C$

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$\int\limits^1_0 {\left( \frac{x}{x^2+1} }\right) \, dx$

substituição
U = x²+1
dU = 2x dx
du/2 = xdx

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