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受欢迎的微积分 >

integral of (ae^θ+b)/(ae^θ-b)

解答

\int \frac{a e ^{θ} + b}{a e ^{θ} - b} d θ

解答

- ln (\frac{e ^{θ} ∣ a ∣}{∣ b ∣}) + 2 ln \frac{a e ^{θ}}{b} - 1 + C

求解步骤

\int \frac{a e ^{θ} + b}{a e ^{θ} - b} d θ

b + a e^{θ} = b (1 + \frac{a e ^{θ}}{b})

= \int \frac{b ( 1 + \frac{a e ^{θ}}{b} )}{- b + a e ^{θ}} d θ

提出常数:

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

= b \cdot \int \frac{1 + \frac{a e ^{θ}}{b}}{- b + a e ^{θ}} d θ

使用换元积分法

= b \cdot \int \frac{u}{b ( u - 1 ) ( u - 2 )} d u

提出常数:

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

= b \frac{1}{b} \cdot \int \frac{u}{( u - 1 ) ( u - 2 )} d u

将

\frac{u}{( u - 1 ) ( u - 2 )}

用部份分式展开:

- \frac{1}{u - 1} + \frac{2}{u - 2}

= b \frac{1}{b} \cdot \int - \frac{1}{u - 1} + \frac{2}{u - 2} d u

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

= b \frac{1}{b} (- \int \frac{1}{u - 1} d u + \int \frac{2}{u - 2} d u)

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{2}{u - 2} d u = 2 ln ∣ u - 2 ∣

= b \frac{1}{b} (- ln ∣ u - 1 ∣ + 2 ln ∣ u - 2 ∣)

u = 1 + \frac{a e ^{θ}}{b}

代回

= b \frac{1}{b} (- ln 1 + \frac{a e ^{θ}}{b} - 1 + 2 ln 1 + \frac{a e ^{θ}}{b} - 2)

化简

b \frac{1}{b} (- ln 1 + \frac{a e ^{θ}}{b} - 1 + 2 ln 1 + \frac{a e ^{θ}}{b} - 2) : - ln (\frac{e ^{θ} ∣ a ∣}{∣ b ∣}) + 2 ln \frac{a e ^{θ}}{b} - 1

= - ln (\frac{e ^{θ} ∣ a ∣}{∣ b ∣}) + 2 ln \frac{a e ^{θ}}{b} - 1

解答补常数

= - ln (\frac{e ^{θ} ∣ a ∣}{∣ b ∣}) + 2 ln \frac{a e ^{θ}}{b} - 1 + C

作图

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