求解这三个积分
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解决时间 2021-12-04 03:40
- 提问者网友：缘字诀
- 2021-12-03 04:04
求解这三个积分
最佳答案
- 五星知识达人网友：刀戟声无边
- 2021-12-03 04:34
1.
∫dx/(x²+2x+3)
=(√2/2)∫1/[1+(x/√2 +1/√2)²]d(x/√2 +1/√2)
=(√2/2)arctan[(x+1)/√2] +C
2.
∫[(x+1)/(x²+2x+3)]dx
=½∫[1/(x²+2x+3)]d(x²+2x+3)
=½ln(x²+2x+3) +C
3.
令x=(3/2)sint，则t=arcsin(⅔x)
∫(1-x)/√(9-4x²)dx
=∫[1-(3/2)sint]/√(9-9sin²t)d[(3/2)sint]
=∫[1-(3/2)sint]·(3/2)cost/(3cost)dt
=¼∫(2-3sint)dt
=¼(2t+3cost) +C
=¼[2arcsin(⅔x) +√(9-4x²)] +C
∫dx/(x²+2x+3)
=(√2/2)∫1/[1+(x/√2 +1/√2)²]d(x/√2 +1/√2)
=(√2/2)arctan[(x+1)/√2] +C
2.
∫[(x+1)/(x²+2x+3)]dx
=½∫[1/(x²+2x+3)]d(x²+2x+3)
=½ln(x²+2x+3) +C
3.
令x=(3/2)sint，则t=arcsin(⅔x)
∫(1-x)/√(9-4x²)dx
=∫[1-(3/2)sint]/√(9-9sin²t)d[(3/2)sint]
=∫[1-(3/2)sint]·(3/2)cost/(3cost)dt
=¼∫(2-3sint)dt
=¼(2t+3cost) +C
=¼[2arcsin(⅔x) +√(9-4x²)] +C
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