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(x^3+1)/(x^2+1)^2的不定积分

如图所示

解:令x^(1/6)=t,则x^(1/3)=t^2,x^(1/2)=t^3,x=t^6,dx=6t^5dt 于是,原式=∫6t^5dt/(t^2+t^3) =6∫t^3dt/(t+1) =6∫[t^2-t+1-1/(t+1)]dt =6(t^3/3-t^2/2+t-ln│t+1│)+C (C是常数) =2t^3-3t^2+6t-6ln│t+1│+C =2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln│x^(...

首先考虑换元法 令x=tant 则dx=(sect)^2 dt 所以原式=∫(sect)^(-3) * (sect)^2 dt =∫(sect)^(-1) dt =∫cost dt =sint + C =tant / √(1+(tant)^2) + C =x/√(1+x^2) + C 完

令x=t^6 则原式=∫6t^5/[t^2(t+1)]dt 把6t^5/[t^2(t+1)]先拆成整式和真分式,再设那个真分式=A/t+B/t^2+C/(t+1),右边通分就得到A、B、C的值,再分别积分。 我就懒得算了……

解题过程: 设x=tant, t=arctanx dx=1/(cost)^2*dt 原式=∫1/√(tan^2t+1)^3*1/cos^2t*dt =∫1/√[(sin^2t+cos^2t)/cos^2t]^3*1/cos^2t*dt =∫cos^3t*1/cos^2t*dt =∫costdt =sint+C =sinarctanx+c 解一些复杂的因式分解问题,常用到换元法,即对结构...

x^3+x-1=x^3+2x-x-1 原式=∫(1/x^2+1)dx-0.5∫(d(x∧2+2)/(x∧2+2)∧2)-∫dx/(x∧2+2)∧2

如图所示:

被积函数 f(x) = (x^3+x−1)/(x^2+2)^2 = (x^3+2x-x−1)/(x^2+2)^2 = x/(x^2+2) - x/(x^2+2)^2 − 1/(x^2+2)^2 I1 = ∫xdx/(x^2+2) = (1/2)∫d(x^2+2)/(x^2+2) = (1/2)ln(x^2+2) + C1, I2 = ∫xdx/(x^2+2)^2 = (1/2)∫d(x^2+2)/(x^2+...

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