原式 =∫1/(b^2)[(a/b)x^2+1)^2]dx
令[√(a/b)]x=tant,t∈(-π/2,π/2)
x=tant[√(b/a)]
原式=[1/(b^2)]∫(cost)^4dt……①
又因为∫(cost)^4dt
=∫[(cos2t+1)/2]^2dt
=(1/4)∫(cos2t)^2dt+(1/4)∫cos2td(2t)
+(1/4)∫dt
=(1/32)∫cos4td(4t)+(3/8)∫dt+(1/4)∫cos2td(2t)
=(1/32)sin4t+(3/8)t+(1/4)sin2t+C…………②
记将②式代回①式后的式子为③式,
将t=arctan[√(a/b)]x代回③式即可.