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sEC²x%1的不定积分

∫dx/(x²+x+1) =4∫dx/(4x²+4x+1+3) =4∫dx/[(2x+1)²+3] = 4/3∫dx/{[(2x+1)/√3]²+1} = 2/√3∫d[(2x+1)/√3]/{[(2x+1)/√3]²+1} =2arctan[(2x+1)/√3]/√3+C

∫ √(x²+1) dx 令x=tanu,则√(x²+1)=secu,dx=sec²udu =∫ sec³u du 下面计算 ∫sec³udu =∫ secudtanu =secutanu - ∫ tan²usecudu =secutanu - ∫ (sec²u-1)secudu =secutanu - ∫ sec³udu + ∫ secudu =secu...

解法一: ∫(x+1)dx =∫(x+1)d(x+1) =½(x+1)²+C 解法二: ∫(x+1)dx =∫xdx+∫dx =½x²+x+C 两个结果形式上看上去不同,其实是积分常数不同造成的。两种解法、两个结果都是正确的。

∫[(x+1)/(x²+1)]dx =∫[x/(x²+1) +1/(x²+1)]dx =(1/2)ln(x²+1) +arctanx +C

一种是换元t=√(x-1)/(2-x)=√(1/(2-x)-1) 得dx=d(2-1/(t²+1))=2t/(t²+1)²dt 然后拆项计算 另一种是分子分母同乘以(2-x) 然后根号内是 √(x-1)(2-x)=√(-x²+3x-2)=√(1/4-(x-3/2)²) 三角换元脱根号令x=3/2+sinu/2后计算

是(x³ +1)/(x² +1)²这个吧,漏了括号,结果都不同

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 +...

4x/(x²+1)(x+1)=(Ax+B)/(x²+1)+C/(x+1) =[(A+C)x² +(A+B)x+B+C]/[(x²+1)(x+1)] A+C=0 A+B=4 B+C=0 A=2 B =2 C=-2 ∫4x/[(x²+1)(x+1)]dx=2∫(x+1)/(x²+1)-1/(x+1)dx =2∫x/(x²+1)+1/(x²+1)-1/(x+1)dx =ln(x...

令x=3sina dx=3cosada √(9-x²)=3cosa a=arcsin(x/3) sina=x/3 则cosa=√(1-x²/9)=√(9-x²)/3 所以原式=∫(1-3sina)/3cosa*3cosada =∫(1-3sina)da =a+3cosa+C =arcsin(x/3)+√(9-x²)/3+C

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