integral of 1/((x^2+10x+24))

解答

\int \frac{1}{( x ^{2} + 10 x + 24 )} d x

\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣ + C

求解步骤

\int \frac{1}{x ^{2} + 10 x + 24} d x

将

\frac{1}{x ^{2} + 10 x + 24}

用部份分式展开:

\frac{1}{2 ( x + 4 )} - \frac{1}{2 ( x + 6 )}

= \int \frac{1}{2 ( x + 4 )} - \frac{1}{2 ( x + 6 )} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 ( x + 4 )} d x - \int \frac{1}{2 ( x + 6 )} d x

\int \frac{1}{2 ( x + 4 )} d x = \frac{1}{2} ln ∣ x + 4 ∣

\int \frac{1}{2 ( x + 6 )} d x = \frac{1}{2} ln ∣ x + 6 ∣

= \frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣

解答补常数

= \frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣ + C

\frac{d}{d x} (ln (arctan (x)))

\frac{d y}{d x} - 3 y = 0

n = 1 \sum \infty (- 0.7)^{n - 1}

\frac{\partial}{\partial t} (\frac{x}{t})

\int_{0}^{3} π (9 x^{2} - x^{4}) d x