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

解答

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

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

求解步骤

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

乘开

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

\int \frac{e ^{3 x}}{e ^{x}} d x = \frac{1}{2} e^{2 x}

\int \frac{e ^{2 x}}{e ^{x}} d x = e^{x}

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

解答补常数

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

\int [u^{5} - u^{3} + u^{2}] d u

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

\int \frac{1}{x ( 3 + x )} d x

\int t sin (t) cos (2 t) d t

\int 18 (x^{2} + 1) (x^{3} + 3 x)^{2} d x