Question

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) dx
3：∫x/√(1-x^2 ) dx
4：∫1/(e^x+1) dx
5：∫lnx/(x(lnx+1)^2 ) dx
麻烦不要直接给答案，稍微有几步步骤，先谢谢了

Answer 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

Answer 2

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 d(1-x^2)=-√((1+x)(1-x))+C

4, ∫1/(e^x+1) dx=∫dx-∫d(1+e^x)/(1+e^x+1) =x-ln(1+e^x )+C

5, ∫lnx/(x(lnx+1)^2 ) dx=∫ lnx/((lnx+1)^2 ) d(lnx)= ∫ d(lnx+1)/((lnx+1)- ∫ d(lnx+1)/(lnx+1)^2= ln(ln(x)+1) + 1/(ln(x)+1)+C