integral of sqrt(t^2+4)

解答

\int t^{2} + 4 d t

4 (\frac{1}{8} t 4 + t^{2} + \frac{1}{2} ln (\frac{t + 4 + t ^{2}}{2})) + C

求解步骤

\int t^{2} + 4 d t

使用三角换元法

= \int 4 sec^{3} (u) d u

提出常数:

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

= 4 \cdot \int sec^{3} (u) d u

使用降次积分法

= 4 (\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) ∣

= 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

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

代回

= 4 (\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

4 (\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))) : 4 (\frac{1}{8} t 4 + t^{2} + \frac{1}{2} ln (\frac{t + 4 + t ^{2}}{2}))

= 4 (\frac{1}{8} t 4 + t^{2} + \frac{1}{2} ln (\frac{t + 4 + t ^{2}}{2}))

解答补常数

= 4 (\frac{1}{8} t 4 + t^{2} + \frac{1}{2} ln (\frac{t + 4 + t ^{2}}{2})) + C

\frac{d}{d x} (sin (x) + \frac{5}{6} cot (x))

\int_{- 8}^{0} (4 - 9 x)^{- 0.5} d x

a re a (x - 1)^{3}, (x - 1)

\frac{\partial}{\partial x} (\frac{1}{2} ln (x^{2} d))

\frac{d}{d x} (csc (x))