What is the integral of 3xe^{3x} ?

The integral of 3xe^{3x} is e^{3x}x-1/3 e^{3x}+C

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integral of 3xe^{3x}

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

e^{3 x} x - \frac{1}{3} e^{3 x} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = 3 x

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

Simplify

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

= e^{3 x} x - \frac{1}{3} e^{3 x}

Add a constant to the solution

= e^{3 x} x - \frac{1}{3} e^{3 x} + C

f (x) = \frac{cos ( t )}{t ^{7}}

x^{^{''}} - 3 x^{^{'}} = e^{- 3 t}

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What is the integral of 3xe^{3x} ?
The integral of 3xe^{3x} is e^{3x}x-1/3 e^{3x}+C