integral of 1/(xsqrt(9+4x^2))

解答

\int \frac{1}{x 9 + 4 x ^{2}} d x

\frac{1}{3} ln tan (\frac{arctan ( \frac{2}{3} x )}{2}) + C

求解步骤

\int \frac{1}{x 9 + 4 x ^{2}} d x

使用三角换元法

= \int \frac{sec ( u )}{3 tan ( u )} d u

提出常数:

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

= \frac{1}{3} \cdot \int \frac{sec ( u )}{tan ( u )} d u

使用换元积分法

= \frac{1}{3} \cdot \int \frac{1}{v} d v

使用积分常见定则:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

代回

= \frac{1}{3} ln tan (\frac{arctan ( \frac{2}{3} x )}{2})

解答补常数

= \frac{1}{3} ln tan (\frac{arctan ( \frac{2}{3} x )}{2}) + C

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0

\frac{d}{d x} (\frac{x - 8}{x + 8})

x \to 4 lim ((x - 4)^{5} - 5)

x \to 3 lim (- 1)

\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{8} x^{\frac{2}{3}} + 5)