What is the integral of 7x*(e^{3x})/7 ?

The integral of 7x*(e^{3x})/7 is 1/9 (3e^{3x}x-e^{3x})+C

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integral of 7x*(e^{3x})/7

\int 7 x \cdot \frac{e ^{3 x}}{7} d x

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

Solution steps

\int 7 x \frac{e ^{3 x}}{7} d x

Take the constant out:

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

= 7 \cdot \frac{1}{7} \cdot \int x e^{3 x} d x

7 \cdot \frac{1}{7} = 1

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

Apply u-substitution

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

= 1 \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 3 x

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

Simplify

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

Add a constant to the solution

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

x \to 0 + lim (x + 2)

3 - sin (2 t) + e^{5 t}

38 (2 e^{x^{2}} x^{2} + e^{x^{2}})

\frac{d y}{d x} = sin (x^{2})

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

What is the integral of 7x*(e^{3x})/7 ?
The integral of 7x*(e^{3x})/7 is 1/9 (3e^{3x}x-e^{3x})+C