integral of sqrt(t^2+2)

解答

\int t^{2} + 2 d t

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

求解步骤

\int t^{2} + 2 d t

使用三角换元法

= \int 2 sec^{3} (u) d u

提出常数:

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

= 2 \cdot \int sec^{3} (u) d u

使用降次积分法

= 2 (\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) ∣

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

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

代回

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

化简

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

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

解答补常数

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

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