integral of sqrt(x^2+21)

解答

\int x^{2} + 21 d x

21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2})) + C

求解步骤

\int x^{2} + 21 d x

使用三角换元法

= \int 21 sec^{3} (u) d u

提出常数:

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

= 21 \cdot \int sec^{3} (u) d u

使用降次积分法

= 21 (\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) ∣

= 21 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{1}{21} x)

代回

= 21 \frac{sec ^{2} ( arctan ( \frac{1}{21} x ) ) sin ( arctan ( \frac{1}{21} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{21} x)) + sec (arctan (\frac{1}{21} x))

化简

21 \frac{sec ^{2} ( arctan ( \frac{1}{21} x ) ) sin ( arctan ( \frac{1}{21} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{21} x)) + sec (arctan (\frac{1}{21} x)) : 21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2}))

= 21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2}))

解答补常数

= 21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2})) + C