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

The integral of (xe^{3x})/3 is 1/27 (3e^{3x}x-e^{3x})+C

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integral of (xe^{3x})/3

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

\frac{1}{27} (3 e^{3 x} x - e^{3 x}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 3 x

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

Simplify

\frac{1}{3} \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) : \frac{1}{27} (3 e^{3 x} x - e^{3 x})

= \frac{1}{27} (3 e^{3 x} x - e^{3 x})

Add a constant to the solution

= \frac{1}{27} (3 e^{3 x} x - e^{3 x}) + C

\int x sin (x) + cos (x) + c d x

\frac{d}{d x} (- 3 cos (x) + x sin (x))

y^{^{'}} (y^{3} - x) = y

\int \frac{1}{x + 9} d x

\frac{d}{d x} (x arcsin (x) + 1 - x^{2})

What is the integral of (xe^{3x})/3 ?
The integral of (xe^{3x})/3 is 1/27 (3e^{3x}x-e^{3x})+C