Question

不定积分计算题

1.x∧2╱（x∧2＋1）∧2的不定积分
2.t^5＋1/t＋1的不定积分
要计算过程
谢谢啦！

Answer 1

(1)
∫x^2/(x^2+1)^2 dx
=∫ dx/(x^2+1) - ∫ dx/(x^2+1)^2
=arctanx - ∫dx/(x^2+1)^2

let
x= tany
dx= (secy)^2 dy
∫dx/(x^2+1)^2
=∫dy/(secy)^2
=∫(cosy)^2 dy
=(1/2)∫(1+cos2y) dy
=(1/2)[y + (1/2)sin2y] + C'
=(1/2)[arctanx + x/(1+x^2)] + C'

∫x^2/(x^2+1)^2 dx
=arctanx - ∫dx/(x^2+1)^2
=arctanx - (1/2)[arctanx + x/(1+x^2)] + C
=(1/2)[arctanx - x/(1+x^2)] + C

(2)
t^5 +1
= t^4(t+1) - t^4 +1
=t^4(t+1) - t^3(t+1) + t^3 +1
=t^4(t+1) - t^3(t+1) + t^2(t+1) -t^2 +1
=t^4(t+1) - t^3(t+1) + t^2(t+1) -t(t+1)+ t+1
=t^4(t+1) - t^3(t+1) + t^2(t+1) -t(t+1)+ (t+1)
ie
t^5 +1 = (t+1)(t^4-t^3+t^2-t+1)
(t^5+1)/(t+1) =t^4-t^3+t^2-t+1

∫ (t^5+1)/(t+1) dt
=∫( t^4-t^3+t^2-t+1) dt
=(1/5)t^5-(1/4)t^4+(1/3)t^3-(1/2)t^2 +t + C

Answer 2