integral of (e^x)/(e^{2x)-3e^x+2}

解答

\int \frac{e ^{x}}{e ^{2 x} - 3 e ^{x} + 2} d x

- ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 2 ∣ + C

求解步骤

\int \frac{e ^{x}}{e ^{2 x} - 3 e ^{x} + 2} d x

将

\frac{e ^{x}}{e ^{2 x} - 3 e ^{x} + 2}

用部份分式展开:

- \frac{1}{e ^{x} - 1} + \frac{2}{e ^{x} - 2}

= \int - \frac{1}{e ^{x} - 1} + \frac{2}{e ^{x} - 2} 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}{e ^{x} - 1} d x + \int \frac{2}{e ^{x} - 2} d x

\int \frac{1}{e ^{x} - 1} d x = ln ∣ e^{x} - 1 ∣ - x

\int \frac{2}{e ^{x} - 2} d x = - x + ln ∣ e^{x} - 2 ∣

= - (ln ∣ e^{x} - 1 ∣ - x) - x + ln ∣ e^{x} - 2 ∣

化简

= - ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 2 ∣

解答补常数

= - ln ∣ e^{x} - 1 ∣ + ln ∣ e^{x} - 2 ∣ + C

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

n = 7 \sum \infty \frac{1}{6 n}

s l o p e y = x^{2} + 5 x, x = 8

\int \frac{3 x ^{3}}{x ^{2} + 1} d x

n = 0 \sum \infty \frac{1}{n}