integral of (e^{4x})/(1+e^x)

解答

\int \frac{e ^{4 x}}{1 + e ^{x}} d x

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

求解步骤

\int \frac{e ^{4 x}}{1 + e ^{x}} d x

使用换元积分法

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

乘开

\frac{( u - 1 ) ^{3}}{u} : u^{2} - 3 u + 3 - \frac{1}{u}

= \int u^{2} - 3 u + 3 - \frac{1}{u} 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

= \int u^{2} d u - \int 3 u d u + \int 3 d u - \int \frac{1}{u} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u d u = \frac{3 u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{3}}{3} - \frac{3 u ^{2}}{2} + 3 u - ln ∣ u ∣

u = 1 + e^{x}

代回

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

解答补常数

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

\int t (1 - ln^{2} (t)) d t

\int (- 2 sin (x) + 4 cos (x)) d x

\int x e^{- (x + y)} d x

\frac{d}{d x} ((7 - 3 x) (x + 5))

y^{^{'}} + 3 y = cos (x)