What is the integral of 9xe^{-9x} ?

The integral of 9xe^{-9x} is -e^{-9x}x-1/9 e^{-9x}+C

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integral of 9xe^{-9x}

\int 9 x e^{- 9 x} d x

- e^{- 9 x} x - \frac{1}{9} e^{- 9 x} + C

Solution steps

\int 9 x e^{- 9 x} d x

Take the constant out:

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

= 9 \cdot \int x e^{- 9 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{81} d u

= 9 \cdot \int \frac{e ^{u} u}{81} d u

Take the constant out:

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

= 9 \cdot \frac{1}{81} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{81} (e^{u} u - \int e^{u} d u)

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

= 9 \cdot \frac{1}{81} (e^{u} u - e^{u})

Substitute back

u = - 9 x

= 9 \cdot \frac{1}{81} (e^{- 9 x} (- 9 x) - e^{- 9 x})

Simplify

9 \cdot \frac{1}{81} (e^{- 9 x} (- 9 x) - e^{- 9 x}) : - e^{- 9 x} x - \frac{1}{9} e^{- 9 x}

= - e^{- 9 x} x - \frac{1}{9} e^{- 9 x}

Add a constant to the solution

= - e^{- 9 x} x - \frac{1}{9} e^{- 9 x} + C

f (x) = x^{2} - 3 x + 3

\frac{d}{d x} (d x)

\frac{1}{s ( s ^{2} + 8 s + 7 )}

\int \frac{1}{1 - 7 x} d x

y^{^{''}} - 2 y^{^{'}} + y = - 2 cos (x) - 2

What is the integral of 9xe^{-9x} ?
The integral of 9xe^{-9x} is -e^{-9x}x-1/9 e^{-9x}+C