integral of sqrt(1+4a^2x^2)

解答

\int 1 + 4 a^{2} x^{2} d x

\frac{1}{2} x 4 a^{2} x^{2} + 1 + \frac{1}{4 a} ln 2 a x + 1 + 4 a^{2} x^{2} + C

求解步骤

\int 1 + 4 a^{2} x^{2} d x

使用三角换元法

= \int \frac{sec ^{3} ( u )}{2 a} d u

提出常数:

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

= \frac{1}{2 a} \cdot \int sec^{3} (u) d u

使用降次积分法

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

u = arctan (2 a x)

代回

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

化简

\frac{1}{2 a} (\frac{sec ^{2} ( arctan ( 2 a x ) ) sin ( arctan ( 2 a x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 a x)) + sec (arctan (2 a x)) ∣) : \frac{1}{2} x 4 a^{2} x^{2} + 1 + \frac{1}{4 a} ln 2 a x + 1 + 4 a^{2} x^{2}

= \frac{1}{2} x 4 a^{2} x^{2} + 1 + \frac{1}{4 a} ln 2 a x + 1 + 4 a^{2} x^{2}

解答补常数

= \frac{1}{2} x 4 a^{2} x^{2} + 1 + \frac{1}{4 a} ln 2 a x + 1 + 4 a^{2} x^{2} + C

\int ln (x - 2) d x

t an g e n t y = \frac{1}{x ^{3}}, (1, 1)

t an g e n t f (x) = \frac{5 x}{x - 3}, (4, 20)

d er i v a t i v e y = (x + 5)^{\frac{1}{x}}

\int x^{2} (x^{3} + 5)^{9} d x