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受欢迎的微积分 >

integral of (3^x-1)/(3^x+1)

解答

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

解答

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

求解步骤

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

使用换元积分法

= \int \frac{u - 1}{ln ( 3 ) u ( u + 1 )} d u

提出常数:

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

= \frac{1}{ln ( 3 )} \cdot \int \frac{u - 1}{u ( u + 1 )} d u

将

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

用部份分式展开:

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

= \frac{1}{ln ( 3 )} \cdot \int - \frac{1}{u} + \frac{2}{u + 1} 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

= \frac{1}{ln ( 3 )} (- \int \frac{1}{u} d u + \int \frac{2}{u + 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

= \frac{1}{ln ( 3 )} (- ln ∣ u ∣ + 2 ln ∣ u + 1 ∣)

u = 3^{x}

代回

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

化简

\frac{1}{ln ( 3 )} (- ln ∣ 3^{x} ∣ + 2 ln ∣ 3^{x} + 1 ∣) : - x + \frac{2}{ln ( 3 )} ln ∣ 3^{x} + 1 ∣

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

解答补常数

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

作图

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