Integrales Definidas

∫▒〖(x^3+x-2)/x^2 dx〗

∫▒(x^3+ x-2)(x^(-2) )dx

∫▒〖(x〗^1 +x^(-1)-2x^(-2))dx

∫▒x^1 dx+∫▒〖x^(-1) dx〗-∫▒〖2x^(-2) dx〗

x^2/2+ln⁡〖IxI+2x^(-1)+c〗

∫▒〖(〖sec〗^2 x)/√tanx dx〗

∫▒〖〖sec〗^2 x*〖tanx〗^(-1/2) dx〗

u=tanx

du=〖sec〗^2 x

∫▒〖u^(-1/2) du〗

2u^(1/2)+c

2〖tan〗^(1/2) x+c

∫▒〖(1+3x)〗^2/∛x dx

∫▒(1^2+(2)(1)(3x)+〖(3x)〗^2)/∛x dx

∫▒〖((1+6x+9x^2))/∛x dx〗

∫▒〖(1+6x+9x^2 )(x^(-1/3))dx〗

∫▒〖1(x^(-1/3))dx〗+∫▒〖〖6x(x〗^(-1/3))dx+ ∫▒〖9x^2 (x^(-1/3))dx〗〗

∫▒〖x^(-1/3) dx〗+6∫▒〖x^(2/3) dx+ 9∫▒〖x^(5/3) dx〗〗

3/2 x^(2/3)+18/5 x^(5/3)+27/8 x^(8/3)+c

∫▒〖〖tan〗^3 (x)dx〗

∫▒tan⁡〖x*〖tan〗^2 x dx〗

∫▒tan⁡〖x*〖(sec〗^2 x-1) dx〗

∫▒〖tan*〖sec〗^2 xdx-∫▒tandx〗

2

Resuelvo las integrales 1 y 2 por separado

Integral 1. ∫▒〖tan*〖sec〗^2 xdx〗

u=tanx

du=〖sec〗^2 x

∫▒udu

1/2 u^2+c

1/2 〖tanx〗^2+c

Integral 2. ∫▒tanxdx

∫▒〖sens/cosx dx〗

u=cox

du=-senxdx -du=sendx

∫▒〖-du/u〗

-lnIuI+c

-lnIcosxI

Ahorra uno las soluciones de las integrales 1 y 2

1/2 tanx^2+lnIxI+c

∫▒√(2+9∛x) /∛(x^2 )

∫▒〖((2+9x^(1/3) )^(1/2))/((x^2 )^(1/3) ) dx〗

∫▒〖(〖(2〗^ +9x^(1/3) )^(1/2))/x^(2/3) dx〗

∫▒〖〖(2〗^ +9x^(1/3) )^(1/2)*x^(-2/3) 〗 dx

∫▒〖〖(2〗^(1/2 )+(9x^(1/3) )^(1/2))*x^(-2/3) dx〗

∫▒〖(2^(1/2)+9x^(1/6) )*x^(-2/3) dx〗

∫▒〖(2^(1/2) x^(2/3)+9x^((-3)/6))dx〗

2^(1/2) ∫▒〖x^(2/3) dx+9∫▒x^((-3)/6) dx〗

2^(1/2) x^(5/3)/(5/3)+9 x^(3/6)/(3/6)+c

〖(2〗^(1/2))(3/5) x^(5/3)+9〖(6/3)x〗^(3/6)+c

6/7 x^(5/3)+18x^(3/6)+c

∫▒〖x/√(3-x^4 ) dx〗

∫▒〖1/√(3-x^4 ) xdx〗

∫▒〖1/√(3-〖(x〗^2 )^2 ) xdx〗

Multiplicamos y dividimos por 3 de radicando

∫▒〖1/(√(3[3-〖(x〗^2 )^2])/3) xdx〗

∫▒〖1/(√3 (3/3)-((x^2 )^2)/3) xdx〗

∫▒〖1/√(3(1-((x^2 )^2)/3)) xdx〗

∫▒〖1/√((1-((x^2 )^2)/3)) x/√3 dx〗

u=x^2/√3

x/√3 dx=1/2 du

∫▒〖1/√(1-u^2 ) 1/2 du〗

1/2 ∫▒〖1/√(1-u^2 ) du〗

1/2 arcsenu

1/2 arcsen x^2/√3

∫▒sen(4x)cos(3x)dx

∫▒〖sen(mx) cos⁡(nx)dx〗

SenA*cosB=1/2(sen(A-B)+(sen(A+B)

∫▒〖1/2((sen(4x-3x)+sen(4x+3x))dx〗

∫▒〖1/2 (senx+sen7x)dx〗

u=x u=7x

du=dx du=7dx

du/7=dx

1/2[∫▒〖sen(u)du+ 1/7 ∫▒〖sen(u)du]〗〗

1/2 ∫▒〖sen(u)du+1/2* 1/7 ∫▒sen(u)du〗

1/2 ∫▒〖sen(u)du+1/14 ∫▒sen(u)du〗

1/2-cosu-1/14 cosu+c

-1/2 cosx-1/14 cos7x+c

∫▒〖1/√(4y^2-4y-3) dy〗

Hallar el valor medio de la función f(x) = x√(x^2 )+16 en el intervalo [0,3]

Aplicamos la fórmula para hallar el valor medio de la función

f(x)= 1/(b-a) ∫_a^b▒〖f(x)〗

1/(3-0) ∫_0^3▒(x√(x^2+16))dx

Hago sustitución y derivo

u=x^2+16

du=2xdx

du/2=xdx

1/3 ∫_0^3▒udu

1/6 ∫_0^3▒〖u^(1/2) du〗

1/6(u^(3/2)/(3/2) )^3+c

1/6(2/3 u^(3/2) )_0^3

2/18

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