integral of sqrt(9+16x^2)

解答

\int 9 + 16 x^{2} d x

\frac{1}{8} (4 x 16 x^{2} + 9 + 9 ln (\frac{4 x + 16 x ^{2} + 9}{3})) + C

求解步骤

\int 9 + 16 x^{2} d x

使用三角换元法

= \int \frac{9}{4} sec^{3} (u) d u

提出常数:

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

= \frac{9}{4} \cdot \int sec^{3} (u) d u

使用降次积分法

= \frac{9}{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) ∣

= \frac{9}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{4}{3} x)

代回

= \frac{9}{4} (\frac{sec ^{2} ( arctan ( \frac{4}{3} x ) ) sin ( arctan ( \frac{4}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{3} x)) + sec (arctan (\frac{4}{3} x)))

化简

\frac{9}{4} (\frac{sec ^{2} ( arctan ( \frac{4}{3} x ) ) sin ( arctan ( \frac{4}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{3} x)) + sec (arctan (\frac{4}{3} x))) : \frac{1}{8} (4 x 16 x^{2} + 9 + 9 ln (\frac{4 x + 16 x ^{2} + 9}{3}))

= \frac{1}{8} (4 x 16 x^{2} + 9 + 9 ln (\frac{4 x + 16 x ^{2} + 9}{3}))

解答补常数

= \frac{1}{8} (4 x 16 x^{2} + 9 + 9 ln (\frac{4 x + 16 x ^{2} + 9}{3})) + C

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