integral of sqrt(9+7x^2)

解答

\int 9 + 7 x^{2} d x

\frac{1}{2} x 7 x^{2} + 9 + \frac{1}{2} \cdot \frac{9}{7} ln (\frac{7 x + 9 + 7 x ^{2}}{3}) + C

求解步骤

\int 9 + 7 x^{2} d x

使用三角换元法

= \int \frac{9}{7} 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}{7} \cdot \int sec^{3} (u) d u

使用降次积分法

= \frac{9}{7} (\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}{7} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{7}{3} x)

代回

= \frac{9}{7} \frac{sec ^{2} ( arctan ( \frac{7}{3} x ) ) sin ( arctan ( \frac{7}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{3} x)) + sec (arctan (\frac{7}{3} x))

化简

\frac{9}{7} \frac{sec ^{2} ( arctan ( \frac{7}{3} x ) ) sin ( arctan ( \frac{7}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{3} x)) + sec (arctan (\frac{7}{3} x)) : \frac{1}{2} x 7 x^{2} + 9 + \frac{1}{2} \cdot \frac{9}{7} ln (\frac{7 x + 9 + 7 x ^{2}}{3})

= \frac{1}{2} x 7 x^{2} + 9 + \frac{1}{2} \cdot \frac{9}{7} ln (\frac{7 x + 9 + 7 x ^{2}}{3})

解答补常数

= \frac{1}{2} x 7 x^{2} + 9 + \frac{1}{2} \cdot \frac{9}{7} ln (\frac{7 x + 9 + 7 x ^{2}}{3}) + C

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