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-x^2 ) dx4:∫1/(e^x+1) dx5:∫lnx/(x(lnx+1)^2 ) dx麻烦不要直接给答案,稍微有几步步骤,
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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-x^2 ) dx4:∫1/(e^x+1) dx5:∫lnx/(x(lnx+1)^2 ) dx麻烦不要直接给答案,稍微有几步步骤,
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,
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-x^2 ) dx4:∫1/(e^x+1) dx5:∫lnx/(x(lnx+1)^2 ) dx麻烦不要直接给答案,稍微有几步步骤,
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...
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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
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