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Pe: 2017-12-03T19:10:53+02:00 In: In:

Integrale

Niste idei…
1.\int \frac{1}{\sqrt{e^{x}+1}}\cdot dx
2.\int \frac{e^{x}(x^{2}-2x+1)}{(x^{2}+1)^{2}}\cdot dx
3.\int_{0}^{1} \frac{x^{2}+1}{x^{4}+1}\cdot dx

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2 raspunsuri

  1. DD
    profesor

    1)Fie ; e^x+1=t^2 -> 2t.dt=e^xdx si e^x=t^2-1->dx=2tdt/(t^2-1)
    I=∫▒1/√((e^x+1) )dx=2∫▒dt/(t^2-1)=∫▒dt/(t-1) – ∫▒dt/(t+1)=ln((t-1)/(t+1))=ln((√((e^x+1) )-1)/(√((e^x+1) )+1))
    2)Vom pune expresia e^x(x^2-2x+1)/(x2+1)^2=e^x/(x^2+1)-e^x.2x/(x^2+1)^2
    I=∫▒〖(e^x.(x^2-2x+1))/(x^2+1)^2 dx〗=∫▒e^x/(x^2+1) dx–∫▒(e^x.(2x))/(((x^2+1)^2 ) ) dx=e^x/(x^2+1)
    3)∫_0^1▒(x^2+1)/(x^4+1)dx=∫_0^1▒(x^2+1)/((x^4+2x^2+1)-2x^2)dx=∫_0^1▒(x^2+1)/((x^2+1)^2-(√2.(x) )^2)dx=
    ∫_0^1▒(x^2+1)/((x^2+√2.x+1).(x^2-√2.x+1) )dx=(.1/2).(∫_0^1▒dx/(x^2-√(2.).x+1)+∫_0^1▒dx/(x^2+√2.x+1))=
    (1/2)((∫_0^1▒dx/((x-1/(√2))^2+1/2))+(∫_0^1▒dx/((x+1/(√2))^2+1/2)) )=√2(arctg(√2.(x-1/(√2)) )-arc tg(√2.(x+1/(√2)) ) )

  2. Integrator
    maestru (V)

    DD post_id=110162 time=1513530001 user_id=13356 wrote:
    1)Fie ; e^x+1=t^2 -> 2t.dt=e^xdx si e^x=t^2-1->dx=2tdt/(t^2-1)
    I=∫▒1/√((e^x+1) )dx=2∫▒dt/(t^2-1)=∫▒dt/(t-1) – ∫▒dt/(t+1)=ln((t-1)/(t+1))=ln((√((e^x+1) )-1)/(√((e^x+1) )+1))
    2)Vom pune expresia e^x(x^2-2x+1)/(x2+1)^2=e^x/(x^2+1)-e^x.2x/(x^2+1)^2
    I=∫▒〖(e^x.(x^2-2x+1))/(x^2+1)^2 dx〗=∫▒e^x/(x^2+1) dx–∫▒(e^x.(2x))/(((x^2+1)^2 ) ) dx=e^x/(x^2+1)
    3)∫_0^1▒(x^2+1)/(x^4+1)dx=∫_0^1▒(x^2+1)/((x^4+2x^2+1)-2x^2)dx=∫_0^1▒(x^2+1)/((x^2+1)^2-(√2.(x) )^2)dx=
    ∫_0^1▒(x^2+1)/((x^2+√2.x+1).(x^2-√2.x+1) )dx=(.1/2).(∫_0^1▒dx/(x^2-√(2.).x+1)+∫_0^1▒dx/(x^2+√2.x+1))=
    (1/2)((∫_0^1▒dx/((x-1/(√2))^2+1/2))+(∫_0^1▒dx/((x+1/(√2))^2+1/2)) )=√2(arctg(√2.(x-1/(√2)) )-arc tg(√2.(x+1/(√2)) ) )


    Bună dimineața,

    Va rog mult , folosiți „tex”….Bieții elevi nu știu ce pot înțelege din raționamentul Dvs.?!?!
    Mulțumesc mult!

    Toate cele bune,

    Integrator

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