integral of (e^x+1)(e^x+x)

解答

\int (e^{x} + 1) (e^{x} + x) d x

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

求解步骤

\int (e^{x} + 1) (e^{x} + x) d x

乘开

(e^{x} + 1) (e^{x} + x) : e^{2 x} + e^{x} x + e^{x} + x

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

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

\int e^{x} x d x = e^{x} x - e^{x}

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

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

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

化简

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

解答补常数

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

d er i v a t i v e x^{4} (2 x - 5)^{6}

y^{^{''}} + 40 y^{^{'}} + 625 y = 0

d er i v a t i v e \frac{s ^{2} + 1}{s ^{2} + 4}

\int_{0}^{1} \frac{17}{x ^{5}} d x

\int x x^{2} + 2 x + 2 d x