integral of sqrt(16x^2+25)

解答

\int 16 x^{2} + 25 d x

\frac{1}{8} (4 x 16 x^{2} + 25 + 25 ln (\frac{4 x + 16 x ^{2} + 25}{5})) + C

求解步骤

\int 16 x^{2} + 25 d x

使用三角换元法

= \int \frac{25}{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{25}{4} \cdot \int sec^{3} (u) d u

使用降次积分法

= \frac{25}{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{25}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{4}{5} x)

代回

= \frac{25}{4} (\frac{sec ^{2} ( arctan ( \frac{4}{5} x ) ) sin ( arctan ( \frac{4}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{5} x)) + sec (arctan (\frac{4}{5} x)))

化简

\frac{25}{4} (\frac{sec ^{2} ( arctan ( \frac{4}{5} x ) ) sin ( arctan ( \frac{4}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{5} x)) + sec (arctan (\frac{4}{5} x))) : \frac{1}{8} (4 x 16 x^{2} + 25 + 25 ln (\frac{4 x + 16 x ^{2} + 25}{5}))

= \frac{1}{8} (4 x 16 x^{2} + 25 + 25 ln (\frac{4 x + 16 x ^{2} + 25}{5}))

解答补常数

= \frac{1}{8} (4 x 16 x^{2} + 25 + 25 ln (\frac{4 x + 16 x ^{2} + 25}{5})) + C

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