integral of 1/(sqrt(4+r^2))

解答

\int \frac{1}{4 + r ^{2}} d r

ln (\frac{1}{2} r + 4 + r^{2}) + C

求解步骤

\int \frac{1}{4 + r ^{2}} d r

使用三角换元法

= \int sec (u) d u

使用积分常见定则:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

u = arctan (\frac{1}{2} r)

代回

= ln tan (arctan (\frac{1}{2} r)) + sec (arctan (\frac{1}{2} r))

化简

ln tan (arctan (\frac{1}{2} r)) + sec (arctan (\frac{1}{2} r)) : ln (\frac{1}{2} r + 4 + r^{2})

= ln (\frac{1}{2} r + 4 + r^{2})

解答补常数

= ln (\frac{1}{2} r + 4 + r^{2}) + C

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