What is the integral of x*ac*e^{-acx+abc-1} ?

The integral of x*ac*e^{-acx+abc-1} is -e^{-acx+acb-1}x-(e^{-acx+acb-1})/(ac)+C

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

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

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

Solution steps

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

Take the constant out:

\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

Apply u-substitution

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

Take the constant out:

\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

Expand

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

Apply the Sum Rule:

\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})

Substitute back

u = - a c x + ab c - 1

= 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})

Simplify

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}

Add a constant to the solution

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

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

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

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

f (x) = x - 5

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

What is the integral of x*ac*e^{-acx+abc-1} ?
The integral of x*ac*e^{-acx+abc-1} is -e^{-acx+acb-1}x-(e^{-acx+acb-1})/(ac)+C