integral of sqrt(t^2+3)

解答

\int t^{2} + 3 d t

3 (\frac{1}{6} t 3 + t^{2} + \frac{1}{2} ln \frac{1}{3} t + \frac{1}{3} (3 + t^{2})) + C

求解步骤

\int t^{2} + 3 d t

使用三角换元法

= \int 3 sec^{3} (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int sec^{3} (u) d u

使用降次积分法

= 3 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= 3 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{1}{3} t)

代回

= 3 \frac{sec ^{2} ( arctan ( \frac{1}{3} t ) ) sin ( arctan ( \frac{1}{3} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} t)) + sec (arctan (\frac{1}{3} t))

化简

3 \frac{sec ^{2} ( arctan ( \frac{1}{3} t ) ) sin ( arctan ( \frac{1}{3} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} t)) + sec (arctan (\frac{1}{3} t)) : 3 (\frac{1}{6} t 3 + t^{2} + \frac{1}{2} ln \frac{1}{3} t + \frac{1}{3} (3 + t^{2}))

= 3 (\frac{1}{6} t 3 + t^{2} + \frac{1}{2} ln \frac{1}{3} t + \frac{1}{3} (3 + t^{2}))

解答补常数

= 3 (\frac{1}{6} t 3 + t^{2} + \frac{1}{2} ln \frac{1}{3} t + \frac{1}{3} (3 + t^{2})) + C

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