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  • ∫x^11(2+3x^4)^(1/2)dx

    ∫x^11(2+3x^4)^(1/2)dx
    數(shù)學(xué)人氣:267 ℃時(shí)間:2020-02-06 00:54:57
    優(yōu)質(zhì)解答
    ∫x^11(2+3x^4)^(1/2)dx=(1/12)∫(2+3x^4)^(1/2)dx^12
    (3x^4/2)=t
    原式=(9/32)∫(2+2t)^(1/2)dt^3=(9/32)t^3(2+2t)^(1/2)-(27√2/(64)∫(t^3+t^2-t^2-t+t)dt/(1+t)^(1/2)
    =(9/32)t^3(2+2t)^(1/2)-(27√2/(64)∫t^2(1+t)^(1/2)dt+(27√2/(64)∫t(1+t)^(1/2)dt-(27√2/(64)∫(t+1)^(1/2)dt+(27√2/(64)∫dt/(1+t)^(1/2)
    (27√2/(64)∫t^2(1+t)^(1/2)dt=(9/64)∫(2+2t)^(1/2)dt^3
    (27/64)∫(2+2t)^(1/2)dt^3=(9/32)t^3(2+2t)^(1/2)+(27√2/128)∫(1+t)^(1/2)dt^2-(9√2/32)(1+t)^(3/2)+(27√2/32)(1+t)^(1/2)=(9/32)t^3(2+2t)^(1/2)+(27√2/128)t^2(1+t)^(1/2)-(27√2/256)∫(t^2+t-t-1+1)dt/(1+t)^(1/2)-(9√2/32)(1+t)^(3/2)+(27√2/32)(1+t)^(1/2)
    =(9/32)t^3(2+2t)^(1/2)+(27√2/128)t^2(1+t)^(1/2)-(27√2/640)(1+t)^(5/2)+(9√2/128)(1+t)^(3/2)+(9√2/128)(1+t)^(3/2)-(27√2/128)(1+t)^(1/2)-(9√2/32)(1+t)^(3/2)+(27√2/32)(1+t)^(1/2)
    =9/32t^3(2+2t)^(1/2)+(27√2/128)t^2(1+t)^(1/2)-(27√2/640)(1+t)^(5/2)-(9√2/64)(1+t)^(3/2)
    原式=(9/32)∫(2+2t)^2dt^3=(27/64)t^3(2+2t)^(1/2)+(81√2/256)t^2(1+t)^(1/2)-(81√2/1280)(1+t)^(5/2)-(27√2/128)(1+t)^(3/2)
    =(27/64)(27/8)x^12(2+3x^4)^(1/2)+(81/256)(9/4)x^8(2+3x^4)^(1/2)-(81√2/(1280)∫(1+3x^4/2)^(5/2)-(27√2/128)(1+3x^4/2)^(3/2)
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