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受欢迎的微积分 >

integral of xace^{-acx+abc-1}

解答

\int x \cdot a c \cdot e^{- a c x + ab c - 1} d x

解答

- e^{- a c x + a c b - 1} x - \frac{e ^{- a c x + a c b - 1}}{a c} + C

求解步骤

\int x a c e^{- a c x + ab c - 1} d x

提出常数:

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

= a c \cdot \int x e^{- a c x + ab c - 1} d x

使用换元积分法

= a c \cdot \int \frac{e ^{u} ( u - a c b + 1 )}{a ^{2} c ^{2}} d u

提出常数:

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

= a c \frac{1}{a ^{2} c ^{2}} \cdot \int e^{u} (u - a c b + 1) d u

乘开

e^{u} (u - a c b + 1) : e^{u} u - a c b e^{u} + e^{u}

= a c \frac{1}{a ^{2} c ^{2}} \cdot \int e^{u} u - a c b e^{u} + e^{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

= a c \frac{1}{a ^{2} c ^{2}} (\int e^{u} u d u - \int a c b e^{u} d u + \int e^{u} d u)

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

\int a c b e^{u} d u = a c b e^{u}

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

= a c \frac{1}{a ^{2} c ^{2}} (e^{u} u - e^{u} - a c b e^{u} + e^{u})

u = - a c x + ab c - 1

代回

= a c \frac{1}{a ^{2} c ^{2}} (e^{- a c x + ab c - 1} (- a c x + ab c - 1) - e^{- a c x + ab c - 1} - a c b e^{- a c x + ab c - 1} + e^{- a c x + ab c - 1})

化简

a c \frac{1}{a ^{2} c ^{2}} (e^{- a c x + ab c - 1} (- a c x + ab c - 1) - e^{- a c x + ab c - 1} - a c b e^{- a c x + ab c - 1} + e^{- a c x + ab c - 1}) : - e^{- a c x + a c b - 1} x - \frac{e ^{- a c x + a c b - 1}}{a c}

= - e^{- a c x + a c b - 1} x - \frac{e ^{- a c x + a c b - 1}}{a c}

解答补常数

= - e^{- a c x + a c b - 1} x - \frac{e ^{- a c x + a c b - 1}}{a c} + C

作图

流行的例子

derivative of 4x^4ln(x)

\frac{d}{d x} (4 x^{4} ln (x))

y^{^{'}} = x - x^{2}

tangent y=x^3-9x,(3,0)

t an g e n t y = x^{3} - 9 x, (3, 0)

f (x) = x - 5

limit as x approaches infinity of-3/x

x \to \infty lim (- \frac{3}{x})