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

解答

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

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

求解步骤

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

使用三角换元法

= \int \frac{sec ( u )}{2 tan ( u )} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{sec ( u )}{tan ( u )} d u

使用换元积分法

= \frac{1}{2} \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}{2} ln ∣ v ∣

代回

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

解答补常数

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

(- 6 x - ln (1 + 6 x) \cdot (1 + 6 x))^{^{'}}

n = 1 \sum \infty \frac{1}{n ^{3} + n ^{2}}

d er i v a t i v e \frac{x ^{2} - 5 x + 3}{x ^{2} + 7}

t an g e n t f (x) = - 5 x^{2} + 2 x

d er i v a t i v e f (x) = 5 x^{2} + 3