Dificuldade em INTEGRAL - atividade estruturada

AJUDA NA RESOLUÇÃO DE INTEGRAIS

Necessito de ajuda para resolver as seguintes integrações:

1) ∫ (2x+1) / √(x²-1) dx

2) ∫ (3x-1)/ (x²+9) dx

3) ∫ (2x + 5)/ (x²+2x+5) dx

4) ∫ (8x - 3)/ √(12x - 4x² - 5) dx

Cálculo I

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ESTÁCIO

veronica galdino

25.11.2013

💡 7 Respostas

Djaldir Alves

25.11.2013

entra nesse site:

http://www.wolframalpha.com

faz um cadastro basico e da pra usar completo por 7 dias. Da a resposta e o passo a passo.

bons estudos

Mario Cesar

26.11.2013

Veronica eu pensei em reslver usando os caracteres de texto, mas acredito que com essas imagens fique facil para você vizualizar a resolução dos exercicios, espero ter ajudado !!!

1)-

$Take the integral:\n integral (2 x+1)\/sqrt(x^2-1) dx\nFor the integrand (2 x+1)\/sqrt(x^2-1), substitute x = sec(u) and dx = tan(u) sec(u) du. Then sqrt(x^2-1) = sqrt(sec^2(u)-1) = tan(u) and u = sec^(-1)(x):\n = integral sec(u) (2 sec(u)+1) du\nExpanding the integrand sec(u) (2 sec(u)+1) gives 2 sec^2(u)+sec(u):\n = integral (2 sec^2(u)+sec(u)) du\nIntegrate the sum term by term and factor out constants:\n = 2 integral sec^2(u) du+ integral sec(u) du\nThe integral of sec^2(u) is tan(u):\n = 2 tan(u)+ integral sec(u) du\nMultiply numerator and denominator of sec(u) by tan(u)+sec(u):\n = 2 tan(u)+ integral (sec^2(u)+sec(u) tan(u))\/(sec(u)+tan(u)) du\nFor the integrand (sec^2(u)+sec(u) tan(u))\/(sec(u)+tan(u)), substitute s = tan(u)+sec(u) and ds = (sec^2(u)+tan(u) sec(u)) du:\n = integral 1\/s ds+2 tan(u)\nThe integral of 1\/s is log(s):\n = log(s)+2 tan(u)+constant\nSubstitute back for s = sec(u)+tan(u):\n = 2 tan(u)+log(tan(u)+sec(u))+constant\nSubstitute back for u = sec^(-1)(x):\nAnswer: | \n | = 2 sqrt(x^2-1)+log(sqrt(x^2-1)+x)+constant$

2)-

$Take the integral:\n integral (3 x-1)\/(x^2+9) dx\nExpanding the integrand (3 x-1)\/(x^2+9) gives (3 x)\/(x^2+9)-1\/(x^2+9):\n = integral ((3 x)\/(x^2+9)-1\/(x^2+9)) dx\nIntegrate the sum term by term and factor out constants:\n = 3 integral x\/(x^2+9) dx- integral 1\/(x^2+9) dx\nFor the integrand x\/(x^2+9), substitute u = x^2+9 and du = 2 x dx:\n = 3\/2 integral 1\/u du- integral 1\/(x^2+9) dx\nFactor 9 from the denominator:\n = 3\/2 integral 1\/u du- integral 1\/(9 (x^2\/9+1)) dx\nFactor out constants:\n = 3\/2 integral 1\/u du-1\/9 integral 1\/(x^2\/9+1) dx\nFor the integrand 1\/(x^2\/9+1), substitute s = x\/3 and ds = 1\/3 dx:\n = 3\/2 integral 1\/u du-1\/3 integral 1\/(s^2+1) ds\nThe integral of 1\/(s^2+1) is tan^(-1)(s):\n = 3\/2 integral 1\/u du-1\/3 tan^(-1)(s)\nThe integral of 1\/u is log(u):\n = (3 log(u))\/2-1\/3 tan^(-1)(s)+constant\nSubstitute back for s = x\/3:\n = (3 log(u))\/2-1\/3 tan^(-1)(x\/3)+constant\nSubstitute back for u = x^2+9:\nAnswer: | \n | = 3\/2 log(x^2+9)-1\/3 tan^(-1)(x\/3)+constant$

3)-

$Take the integral:\n integral (2 x+5)\/(x^2+2 x+5) dx\nRewrite the integrand (2 x+5)\/(x^2+2 x+5) as (2 x+2)\/(x^2+2 x+5)+3\/(x^2+2 x+5):\n = integral ((2 x+2)\/(x^2+2 x+5)+3\/(x^2+2 x+5)) dx\nIntegrate the sum term by term and factor out constants:\n = 3 integral 1\/(x^2+2 x+5) dx+ integral (2 x+2)\/(x^2+2 x+5) dx\nFor the integrand (2 x+2)\/(x^2+2 x+5), substitute u = x^2+2 x+5 and du = (2 x+2) dx:\n = integral 1\/u du+3 integral 1\/(x^2+2 x+5) dx\nFor the integrand 1\/(x^2+2 x+5), complete the square:\n = integral 1\/u du+3 integral 1\/((x+1)^2+4) dx\nFor the integrand 1\/((x+1)^2+4), substitute s = x+1 and ds = dx:\n = 3 integral 1\/(s^2+4) ds+ integral 1\/u du\nFactor 4 from the denominator:\n = 3 integral 1\/(4 (s^2\/4+1)) ds+ integral 1\/u du\nFactor out constants:\n = 3\/4 integral 1\/(s^2\/4+1) ds+ integral 1\/u du\nFor the integrand 1\/(s^2\/4+1), substitute p = s\/2 and dp = 1\/2 ds:\n = 3\/2 integral 1\/(p^2+1) dp+ integral 1\/u du\nThe integral of 1\/(p^2+1) is tan^(-1)(p):\n = 3\/2 tan^(-1)(p)+ integral 1\/u du\nThe integral of 1\/u is log(u):\n = 3\/2 tan^(-1)(p)+log(u)+constant\nSubstitute back for p = s\/2:\n = 3\/2 tan^(-1)(s\/2)+log(u)+constant\nSubstitute back for s = x+1:\n = log(u)+3\/2 tan^(-1)((x+1)\/2)+constant\nSubstitute back for u = x^2+2 x+5:\nAnswer: | \n | = log(x^2+2 x+5)+3\/2 tan^(-1)((x+1)\/2)+constant$

4)-

$Take the integral:\n integral (8 x-3)\/sqrt(-4 x^2+12 x-5) dx\nFor the integrand (8 x-3)\/sqrt(-4 x^2+12 x-5), complete the square:\n = integral (8 x-3)\/sqrt(4-(2 x-3)^2) dx\nFor the integrand (8 x-3)\/sqrt(4-(2 x-3)^2), substitute u = 2 x-3 and du = 2 dx:\n = 1\/2 integral (4 u+9)\/sqrt(4-u^2) du\nFor the integrand (4 u+9)\/sqrt(4-u^2), substitute u = 2 sin(s) and du = 2 cos(s) ds. Then sqrt(4-u^2) = sqrt(4-4 sin^2(s)) = 2 cos(s) and s = sin^(-1)(u\/2):\n = 1\/2 integral (8 sin(s)+9) ds\nIntegrate the sum term by term and factor out constants:\n = 9\/2 integral 1 ds+4 integral sin(s) ds\nThe integral of 1 is s:\n = (9 s)\/2+4 integral sin(s) ds\nThe integral of sin(s) is -cos(s):\n = (9 s)\/2-4 cos(s)+constant\nSubstitute back for s = sin^(-1)(u\/2):\n = 9\/2 sin^(-1)(u\/2)-2 sqrt(4-u^2)+constant\nSubstitute back for u = 2 x-3:\nAnswer: | \n | = -2 sqrt(-4 x^2+12 x-5)-9\/2 sin^(-1)(3\/2-x)+constant$