高数不定积分 求解答啊 急!!!

1、∫sin(lnx)dx =xsin(lnx)-∫xcos(lnx)*(1/x)dx =xsin(lnx)-∫cos(lnx)dx =xsin(lnx)-xcos(lnx)-∫xsin(lnx)(1/x)dx =xsin(lnx)-xcos(lnx)-∫sin(lnx)dx 将-∫sin(lnx)dx移到左边与左边合并，除去系数得：∫sin(lnx)dx=1/2xsin(lnx)-1/2xcos(lnx)+C 2、∫cos(lnx...

答：设1+e^x=t，x=ln(t-1)∫ (1+e^x)^(-1) dx =∫ 1/(1+e^x) dx =∫ 1/t d [ln(t-1)]=∫ (1/t)*1/(t-1) dt =∫ 1/(t-1) -1/t dt =ln(t-1)-lnt+C =x-ln(1+e^x)+C

【俊狼猎英】团队为您解答 1）de^x=e^xdx 原积分=∫2√[1-(e^x)^2]de^x 令e^x=sint 原积分=2∫(cost)^2dt =∫cos2t-1dt =sin2t/2-t+C =e^x√(1-e^2x)-arcsin(e^x)+C 2）把分子拆为(x-3)+3 原积分=-∫dx/(3-x)^6+3∫dx/(3-x)^7 =1/5(3-x)^5-1/2...

∫cscx dx =∫1/sinx dx =∫1/[2sin(x/2)cos(x/2)] dx,两倍角公式 =∫1/[sin(x/2)cos(x/2)] d(x/2)=∫1/tan(x/2)*sec²(x/2) d(x/2)=∫1/tan(x/2) d[tan(x/2)],注∫sec²(x/2)d(x/2)=tan(x/2)+C =ln|tan(x/2)|+C,这是答案一 进一...

2.分部积分：∫ln(1+x^2)dx =xln(1+x^2)-∫2x^2/(1+x^2)dx =xln(1+x^2)-[∫2dx-∫2/(1+x^2)dx]=xln(1+x^2)-2x+2arctanx+C 3.∫(xcos2x)dx =(1/2)∫xdsin2x =(1/2)xsin2x-(1/2)∫sin2xdx =(1/2)xsin2x-(1/4)∫sin2xd2x =(1/2)xsin2x+(1/4)...

sect]=∫dt/[cost((2sin²t/cos²t)+1)]=∫costdt/[((2sin²t+cost²)]=∫[1/(1+sin²t)]d(sint)=arctan(sint)+C 三角替换有sint=x/√(1+x²)所以原不定积分 ∫dx/(2x^2+1)(x^2+1)^(1/2)=arctan[x/√(1+x²)]+C ...

=∫xsec²xdx-∫xdx =∫xdtanx-x²/2 =xtanx-∫tanxdx-x²/2 =xtanx-∫sinx/cosx dx-x²/2 =xtanx-∫-dcosx/cosx-x²/2 =xtanx+ln|cosx|-x²/2+C 11、∫cos(lnx)dx =xcos(lnx)-∫xdcos(lnx)=xcos(lnx)-∫x[-sin(lnx)]*(lnx)'dx =...

由已知条件 {F(x)^2}'=xe^(x^2)积分求出 F(x)^2, 然后算出F(x)然后求导就可以算出f(x)

=∫(-1->1) x[ e^x -e^(-x) ]dx =2∫(0->1) x[ e^x -e^(-x) ]dx =2( e + 1/e)∫(0->1) xe^x dx =∫(0->1) xde^x =[xe^x]|(0->1) -∫(0->1) e^x dx =e - [e^x]|(0->1)=1 ∫(0->1) xe^(-x) dx =-∫(0->1) xde^(-x)=-[...

回答：原式=∫2dx/√(9-4x^2)-∫xdx/√(9-4x^2) =∫d(2x/3)/√[1-(2x/3)^2]+1/8∫d(9-4x^2)/√(9-4x^2) =arcsin(2x/3)+(1/4)√(9-4x^2)+C