哪位能帮忙做下这几道积分题1：∫1/(x^2+a^2 ) dx (a ≠0)2：∫x√(x-1) dx3：∫x/√(1-

1.由∫1//(x^2+1)dx=arctanx,令x/a=t,则式=(1/a).arctan(x/a)+C
2.令t=√(x-1),则dx=2tdt;则式=∫(t^2-1)t*2tdt=2∫(t^4-t^2)dt=2t^3(t^2/5-1/3)+c;
so =2/5√(x-1)^5- 2/3*√(x-1)^3+C;
3.用到换元-0.5∫1/√(1-x^2 )d(1-x^2),再换
4.=∫(1+e^x-e^x)/(e^x+1) dx =∫[1-e^x/(e^x+1)] dx = x-ln(e^x+1)+C
5.∫lnx/(x(lnx+1)^2 ) dx,令lnx=t,so dx=(e^t)*dt,so =∫t*e^t/(e^t*(t+1)^2 ) dt=
∫t/(t+1)^2 dx= ∫[1/(t+1)-1/(t+1)^2]dt= ln(lnx+1)+1/(lnx+1)+C
记不大住了,就这样吧,不好意思了

1, 代入x'=x/a, 用公式d/dx tan^-1(x)=1/(1+x^2),
∫1/(x^2+a^2 ) dx =(1/a) (tan^-1(x/a))+C
2, x√(x-1)=(x-1)^1.5+√(x-1)
∫x√(x-1) dx=(1/3)(2x+4)√(x-1)+C
3, ∫x/√(1-x^2 ) dx=(-1/2)∫(1-x^2 )^-0.5...