integral of 2/(sqrt(x^2+9))

解答

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

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

求解步骤

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

提出常数:

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

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

使用三角换元法

= 2 \cdot \int sec (u) d u

使用积分常见定则:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 2 ln ∣ tan (u) + sec (u) ∣

u = arctan (\frac{1}{3} x)

代回

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

化简

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

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

解答补常数

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

n = 0 \sum \infty \frac{n}{1 + n ^{2}}

t an g e n t f (x) = \frac{2}{x ^{2}}, (7, \frac{2}{49})

\frac{d}{d x} (ln (3 x^{2} + 1))

\frac{d}{d y} (\frac{y}{y + \frac{b}{y}})

n = 1 \sum \infty - \frac{1}{3 ( 3 n + 2 )}