Question

求教这题怎么积分

Answer 1

采用公式或用分部积分

Answer 2

令 x = atanu， 则
I = ∫a(secu)^2du/(asecu)^4 = (1/a^3)∫(cosu)^2du
= [1/(2a^3)]∫(1+cos2u)du
= [1/(2a^3)][u+(1/2)sin2u] + C
= [1/(2a^3)][u+sinucosu] + C
= [1/(2a^3)][arctan(x/a)+ax/(x^2+a^2)] + C

Answer 3

代入x=atanu即可解决问题:dx=asec²udu
(x²+a²)²=(a²+a²tan²u)²=(a²sec²u)²=a⁴sec⁴u
即(x²+a²)²=a⁴/cos⁴u
cos⁴u=a⁴/(x²+a²)²→cosu=a/√(x²+a²)
sinu=x/√(x²+a²)
∴∫dx/(a²+x²)²
=∫asec²u/(a⁴sec⁴u) du
=(1/a³)∫cos²u du
=(1/2a³)∫(1+cos2u) du
=(1/2a³)(u+1/2*sin2u)+C
=(1/2a³)(u+sinu*cosu)+C
=(1/2a³)[arctan(x/a)+x/√(x²+a²)*a/√(x²+a²)]+C
=(1/2a³)[arctan(x/a)+ax/(x²+a²)]+C