What is the integral of x*ab*e^{-abx+abc} ?

The integral of x*ab*e^{-abx+abc} is -e^{-abx+abc}x-(e^{-abx+abc})/(ab)+C

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integral of xabe^{-abx+abc}

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

- e^{- ab x + ab c} x - \frac{e ^{- ab x + ab c}}{ab} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

e^{u} (u - ab c) : e^{u} u - ab c e^{u}

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

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

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

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

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

Substitute back

u = - ab x + ab c

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

Simplify

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

= - e^{- ab x + ab c} x - \frac{e ^{- ab x + ab c}}{ab}

Add a constant to the solution

= - e^{- ab x + ab c} x - \frac{e ^{- ab x + ab c}}{ab} + C

What is the integral of x*ab*e^{-abx+abc} ?
The integral of x*ab*e^{-abx+abc} is -e^{-abx+abc}x-(e^{-abx+abc})/(ab)+C