Question

高数上，求积分问题

积分{（x^3+1）/[(x^2+1)^2]}dx

给下解题步骤，最好带下思路，感激不尽！

Answer 1

分子上加x减x，拆为两个积分，后一个积分再拆为两个，前两个很简单，后一个稍烦些，
∫（x^3+1）/[(x^2+1)^2] dx
=∫（x^3+1+x-x）/[(x^2+1)^2] dx
=∫（x^3+x）/[(x^2+1)^2] dx-∫x/[(x^2+1)^2] dx+∫1/[(x^2+1)^2] dx
=∫ x/(x^2+1)dx-∫x/[(x^2+1)^2] dx+∫1/[(x^2+1)^2] dx
=(1/2)∫ 1/(x^2+1)d(x^2)-(1/2)∫1/[(x^2+1)^2] d(x^2)+∫1/[(x^2+1)^2] dx
=(1/2)ln(x^2+1)+(1/2)1/(x^2+1)+∫1/[(x^2+1)^2] dx
第三个积分计算如下
x=tanθ，dx=sec²θdθ，sinθ=x/√(1+x²)，cosθ=1/√(1+x²)
∫1/(1+x²)² dx
= ∫1/sec⁴θ * sec²θ dθ
= ∫cos²θ dθ
= (1/2)∫(1+cos2θ) dθ
= (1/2)θ + (1/2)sinθcosθ + C
= (1/2)[arctanx + x/(1+x²)] + C

有不明白请追问

Answer 2

分母加x减x，裂项然后分项积分即可
原式=∫x/(x²+1)dx-∫x/(x²+1)²dx+∫1/(x²+1)²dx
主要解出∫/(x²+)²dx即可
因为
∫1/(x²+1)dx=x/(x²+1)+∫2x²/(x²+1)²dx
=x/(x²+1)+2∫1/(x²+1)dx-2∫1/(x²+1)²dx
即可解出