What is the integral of 63xe^{3x} ?

The integral of 63xe^{3x} is 7(3e^{3x}x-e^{3x})+C

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integral of 63xe^{3x}

\int 63 x e^{3 x} d x

7 (3 e^{3 x} x - e^{3 x}) + C

Solution steps

\int 63 x e^{3 x} d x

Take the constant out:

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

= 63 \cdot \int x e^{3 x} d x

Apply u-substitution

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

= 63 \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

= 63 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 63 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 63 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 x

= 63 \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

Simplify

63 \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) : 7 (3 e^{3 x} x - e^{3 x})

= 7 (3 e^{3 x} x - e^{3 x})

Add a constant to the solution

= 7 (3 e^{3 x} x - e^{3 x}) + C

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What is the integral of 63xe^{3x} ?
The integral of 63xe^{3x} is 7(3e^{3x}x-e^{3x})+C