What is the integral of xe^{-5x} ?

The integral of xe^{-5x} is 1/25 (-5e^{-5x}x-e^{-5x})+C

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integral of xe^{-5x}

\int x e^{- 5 x} d x

\frac{1}{25} (- 5 e^{- 5 x} x - e^{- 5 x}) + C

Solution steps

\int x e^{- 5 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{25} d u

= \int \frac{e ^{u} u}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{25} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{25} (e^{u} u - e^{u})

Substitute back

u = - 5 x

= \frac{1}{25} (e^{- 5 x} (- 5 x) - e^{- 5 x})

Simplify

= \frac{1}{25} (- 5 e^{- 5 x} x - e^{- 5 x})

Add a constant to the solution

= \frac{1}{25} (- 5 e^{- 5 x} x - e^{- 5 x}) + C

\int (tan (θ) + cot (θ))^{2} d θ

\int \frac{6}{x} + 6 x d x

(- 27 sin (3 x) - 27 x cos (3 x))^{^{'}}

f (x) = \frac{2 x}{x ^{2} + 1}, at x = - 1, - 1

\frac{2 e ^{- s}}{( s + \frac{1}{5} )}

What is the integral of xe^{-5x} ?
The integral of xe^{-5x} is 1/25 (-5e^{-5x}x-e^{-5x})+C