作业帮求答,高数积分两条
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求答,高数积分两条
求答,高数积分两条
求答,高数积分两条
Let x = tanθ and dx = sec²θ dθ
∫ dx/(x²+1)^(3/2)
= ∫ (sec²θ)/(tan²θ+1)^(3/2) dθ
= ∫ (sec²θ)/(sec²θ)^(3/2) dθ
= ∫ (sec²θ)/(sec³θ) dθ
= ∫ cosθ dθ= sinθ + C= x/√(1+x²) + C