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受欢迎的微积分 >

integral of sqrt(9+t^2)

解答

\int 9 + t^{2} d t

解答

9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3})) + C

求解步骤

\int 9 + t^{2} d t

使用三角换元法

= \int 9 sec^{3} (u) d u

提出常数:

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

= 9 \cdot \int sec^{3} (u) d u

使用降次积分法

= 9 (\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) ∣

= 9 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

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

代回

= 9 (\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

9 (\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))) : 9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3}))

= 9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3}))

解答补常数

= 9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3})) + C

作图

流行的例子

(dy)/(dx)= y/x+(9x^2)/(y^2)

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