integral from 0 to 1 of (ax)/(x^2-px)

解答

\int_{0}^{1} \frac{a x}{x ^{2} - p x} d x

a (ln ∣ 1 - p ∣ - ln ∣ p ∣)

求解步骤

\int_{0}^{1} \frac{a x}{x ^{2} - p x} d x

提出常数:

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

= a \cdot \int_{0}^{1} \frac{x}{x ^{2} - p x} d x

化简

\frac{x}{x ^{2} - p x} : \frac{1}{x - p}

= a \cdot \int_{0}^{1} \frac{1}{x - p} d x

使用换元积分法

= a \cdot \int_{- p}^{1 - p} \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= a [ln ∣ u ∣]_{- p}^{1 - p}

带入上下限计算后得:

ln ∣ 1 - p ∣ - ln ∣ p ∣

= a (ln ∣ 1 - p ∣ - ln ∣ p ∣)

\int 3 x^{2} + 8 x - 2 d x

\frac{\partial}{\partial t} (t^{4} cos (s))

d er i v a t i v e csc^{3} (5 x^{2} + 1)

y^{^{'}} + \frac{2 y}{x} = - \frac{2}{x}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 0