integral of 1/(3^x+1)

解答

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

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

求解步骤

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

\frac{1}{3 ^{x} + 1} = - \frac{3 ^{x}}{3 ^{x} + 1} + 1

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

\int \frac{3 ^{x}}{3 ^{x} + 1} d x = \frac{1}{ln ( 3 )} ln ∣ 3^{x} + 1 ∣

\int 1 d x = x

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

解答补常数

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 3, y^{^{'}} (0) = 5

\int_{1}^{3} (x + 2) (x - 1)^{5} d x

\frac{d}{d x} (\frac{- x ^{2} - 4 x - 1}{4 x - 3})

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

\frac{d}{d x} (ln (e^{3 x} + e^{- 3 x}))