x2-1的积分

熊虹伦5094(x^2+2x - 1)/(x - 1)(x^2 - x+1)的不定积分 -
鲍哄维18374136575 ______[答案] (x^2+2x-1)/(x-1)(x^2-x+1) =2/(x-1)+(-x+3)/(x^2-x+1) =2/(x-1)-(1/2)(2x-1-5)/(x^2-x+1) =2/(x-1)-(1/2)(2x-1)/(x^2-x+1)-(5/2)/(x^2-x+1) 所以:∫(x^2+2x-1)/(x-1)(x^2-x+1)dx ∫2/(x-1)dx-(1/2)∫(2x-1)/(x^2-x+1)dx-(5/2)∫1/(x^2-x+1)dx =2ln|x-1|-(1/2)ln(x^2-x+1)-(5/√3)...

熊虹伦5094f(x)=(2x - 1)/(x - 1),求该函数在下限0上限e+1上的积分
鲍哄维18374136575 ______ 先化简,得(2x-1)/(x-1)=(2x-2)/(x-1)+1/(x-1)=1+1/(x-1) 所以F (x)=x+ln(x-1) 最后 积分=e+1+ln(e+1-1)=e+1+1=e+2

熊虹伦50941/1 - x2 不定积分 -
鲍哄维18374136575 ______[答案] ∫ dx/(1-x²)令1/(1-x²) = A/(1+x) + B/(1-x)则1 = A(1-x) + B(1+x)当x=-1,1=2A => A=1/2当x=1,1=2B => B=1/2∴原式= (1/2)∫ dx/(1+x) + ∫ dx/(1-x)= (1/2)∫ d(1+x)/(1+x) ∫ -d(1-x)/(1-x)= (1/2)ln|1...

熊虹伦5094求∫(x2/(√(1 - x2)dx的不定积分 -
鲍哄维18374136575 ______[答案] 设x=sint,t=arcsinx,dx=costdt, 原式=∫(sint)^2*costdt/cost =∫(sint)^2dt =(1/2)∫(1-cos2t)dt =t/2-(1/4)sin2t+C =(arcsinx)/2-(1/2)x√(1-x^2)+C.

熊虹伦5094【根号下(e^2x - 1)】分之一的积分 -
鲍哄维18374136575 ______[答案] 设e^x=sect=1/cost,则x=lnsect,t=arcsec(e^x)=arccos(1/e^x)则dx=(1/sect)·(sint/cos²t)dt=tantdte^2x-1=sec²t-1=tan²t所以原式=∫1/√(e^2x-1)dx=∫(1/tant)tantdt=t+C=arccos(1/e^x)+C

熊虹伦5094根号下(e^2x) - 1分之一的积分 -
鲍哄维18374136575 ______[答案] 见下图:

熊虹伦50941/x²√2x - 1的不定积分 -
鲍哄维18374136575 ______ 展开全部(x^2+2x-1)/(x-1)(x^2-x+1)=2/(x-1)+(-x+3)/(x^2-x+1)=2/(x-1)-(1/2)(2x-1-5)/(x^2-x+1)=2/(x-1)-(1/2)(2x-1)/(x^2-x+1)-(5/2)/(x^2-x+1) 所以:∫(x^2+2x-1)/(x-1)(x^2-x+1)dx ∫2/(x-1)dx-(1/2)∫(2x-1)/(x^2-x+1)dx-(5/2)∫1/(x^2-x+1)dx=2ln|x-1|-(1/2)ln(x^2-x+1)-(5/√3)arctan((2x-1)/√3)+C

熊虹伦5094不定积分(x2 - a2)的1/2次方除以x的解题过程 -
鲍哄维18374136575 ______[答案] 换元,设x=asect,则dx=asect*tgtdt 原式=a*积分(tgt)^2dt=a*积分[(sect)^2-1]dt=atgt-at+c=(x2-a2)的1/2次方-a*arccos(a/x)+c,c为常数.

熊虹伦5094根号(1 - x2/1+x2)的不定积分 -
鲍哄维18374136575 ______[答案] 用分部积分: ∫ √(x^2 +1)dx=x√(x^2 +1)-∫ x^2dx/√(x^2 +1) =x√(x^2 +1)-∫ (x^2+1-1)dx/√(x^2 +1) =x√(x^2 +1)-∫ √(x^2+1)dx+∫ dx/√(x^2 +1) =x√(x^2 +1)+ln[x+√(x^2 +1)-∫ √(x^2 +1)dx 移项:除以2 ∫√(x^2 +1)dx=(x/2)√(x^2 +1)+(1/2)ln[x+√(x^2 +1)+C

熊虹伦50941/(1 - x2)2的积分怎么求啊?两个2都是指数! -
鲍哄维18374136575 ______[答案] 用代换法.将1-x2=t 然后求出1/t2的积分.再把dx换算成dt然后就求出来了