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∫Dx/{[(x+1)^2][(x%1)^4]}^(1/3)的不定积分积分

解:设(x-1)/(x+1)=t³,则dx=6t²dt/(1-t³)²,x+1=2/(1-t³),x-1=2t³/(1-t³) ∴原式=∫[6t²dt/(1-t³)²]/{[2/(1-t³)]^(2/3)*[2t³/(1-t³)]^(4/3)} =∫(6t²dt)/[2²t^4] =3/2...

这样换换元后,无理函数的积分就转化为有理函数的积分。

∫[1/(1+x^4)]dx = 1/2∫[(x^2+1)-(x^2-1)]/(1+x^4)dx = 1/2 {∫(x^2+1)/(1+x^4) dx - ∫(x^2-1)/(1+x^4)dx } = 1/2 {∫(1+1/x^2)dx /(x^2+1/x^2) - ∫(1-1/x^2)dx/(x^2+1/x^2)} = 1/2 {∫d(x-1/x) /[(x-1/x)^2+2] - ∫d(x+1/x) /[(x+1/x)^2 -...

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^(...

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……

=∫½[1+cos(2x)]dx =∫½dx+∫½cos(2x)dx =∫½dx+¼∫cos(2x)d(2x) =½x+¼sin(2x) +C 解题思路: 先运用二倍角公式进行化简。 cos(2x)=2cos²x-1 则cos²x=½[1+cos(2x)]

可以先把x+2提出来 1/[(x-1)^(1/3)*(x+2)^(2/3)] =1/(x+2) * [(x+2)/(x-1)]^(1/3) 做换元,令t=[(x+2)/(x-1)]^(1/3) 则x=1 + 3/(t^3-1) dx=9t^2dt/(t^3-1)^2 代入原式得 ∫3 / [t(t^3-1)] dt =∫ -3/t + 1/(t-1) + (2t+1)/(t^2+t+1) dt =-3ln|t| +...

先分解因式: ∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx = ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx 1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C 1 = (A + B)x² +...

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