What is the integral of-x*e^{3x} ?

The integral of-x*e^{3x} is -1/9 (e^{3x}*3x-e^{3x})+C

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integral of-x*e^{3x}

\int - x \cdot e^{3 x} d x

- \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + C

Solution steps

\int - x e^{3 x} d x

Take the constant out:

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

= - \int x e^{3 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= - \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 3 x

= - \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

Add a constant to the solution

= - \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + C

\int x^{2} - y^{2} d y

\int \frac{4 x ^{3} - 6 x ^{4}}{2 x ^{6}} d x

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

(\frac{x ^{2} + x + 4}{x + 1})^{^{'}}

8 t e^{2 t}

What is the integral of-x*e^{3x} ?
The integral of-x*e^{3x} is -1/9 (e^{3x}*3x-e^{3x})+C