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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...

思路应该是:提出(x-1)(x+1)之后,对其余部分的替换。 具体过程如下:被积函数 ³√(x+1)²(x-1)(x-1)³ =(x-1) ³√(x+1)²(x-1) =(x-1) ³√(x+1)³(x-1)/(x+1) =(x-1)(x+1) ³√(x-1)/(x+1)

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……

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| +...

x=tant ∫dx/[x⁴√(1+x²)]=∫dtant/[tan⁴t√(1+tan²t)] = ∫sect/tan⁴tdsint=∫cos³t/sin⁴tdt =∫cos²t/sin⁴tdsint=∫1 /sin⁴ t-1/sin⁴tdsint =-1/sint+1/(3sin³t)+C =-sect/tant+s...

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 +...

45 令 x^(1/4)=u, 则 x=u^4, dx=4u^3du ∫dx/[x^(1/2)+x^(1/4)] = ∫4u^3du/[u^2+u] = 4∫[u-1+1/(u+1)]du = 2u^2-4u+4ln|u+1|+C = 2x^(1/2)-4x^(1/4)+4ln[x^(1/4)+1]+C

分解部分分式时,分母为二次式的项其分子的一般形式就是 ax+b这样 一次二项式 ,一般不能预先知道a,b会不会有哪个为0, 所以,你那个题,拆成 a/(x-1)+(bx+c)/(x∧2+1)+d/(x+1)再来确定系数才是常规方法。 事实上:

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