∫ 1/(9+4x^2)dx=?

题目

∫ 1/(9+4x^2)dx=?

答案

∫ 1/(9 + 4x²) dx
= ∫ 1/[4(9/4 + x²)] dx
= (1/4)∫ 1/[(3/2)² + x²] dx
= (1/4) * 1/(3/2) * arctan[x/(3/2)] + C
= (1/6)arctan(2x/3) + C
或
令2x = 3tanz,z = arctan(2x/3),2 dx = 3 sec²z dz
∫ 1/(9 + 4x²) dx = ∫ 1/(9 + 9tan²z) (3/2 * sec²z dz)
= ∫ 1/(9sec²z) * (3/2 sec²z dz)
= (1/6)∫ dz = z/6 + C
= (1/6)arctan(2x/3) + C