∫ 1/(9+4x^2)dx=?
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解决时间 2021-04-27 11:20
- 提问者网友:寂寞撕碎了回忆
- 2021-04-27 05:20
∫ 1/(9+4x^2)dx=?
最佳答案
- 五星知识达人网友:逐風
- 2021-04-27 06:17
∫ 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
= ∫ 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
全部回答
- 1楼网友:摆渡翁
- 2021-04-27 07:52
1/2*1/3arttan2x/3 c追问额 不是吧?追答哦,打错了,是1/6arctan2x/3 +c
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