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求不定积分∫[1/(x^2+4)]dx
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求不定积分∫[1/(x^2+4)]dx
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答案和解析
方法一：运用公式∫ dx/(a² + b²x²) = (1/ab)arctan(bx/a) + C
∫ dx/(x² + 4) = (1/2)arctan(x/2) + C
方法二：三角函数换元法：令x = 2tanz,dx = 2sec²z dz
∫ dx/(x² + 4)
= ∫ (2sec²z dz)/(4tan²z + 4)
= ∫ 2sec²z/[4(tan²z + 1)] dz
= (1/2)∫ sec²z/sec²z dz
= z/2 + C
= (1/2)arctan(x/2) + C,因为tanz = x/2
∫ dx/(x² + 4) = (1/2)arctan(x/2) + C
方法二：三角函数换元法：令x = 2tanz,dx = 2sec²z dz
∫ dx/(x² + 4)
= ∫ (2sec²z dz)/(4tan²z + 4)
= ∫ 2sec²z/[4(tan²z + 1)] dz
= (1/2)∫ sec²z/sec²z dz
= z/2 + C
= (1/2)arctan(x/2) + C,因为tanz = x/2
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