What is the integral of 3xe^{7x} ?

The integral of 3xe^{7x} is 3/49 (7e^{7x}x-e^{7x})+C

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integral of 3xe^{7x}

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

\frac{3}{49} (7 e^{7 x} x - e^{7 x}) + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{49} d u

= 3 \cdot \int \frac{e ^{u} u}{49} d u

Take the constant out:

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

= 3 \cdot \frac{1}{49} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{49} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{49} (e^{u} u - e^{u})

Substitute back

u = 7 x

= 3 \cdot \frac{1}{49} (e^{7 x} \cdot 7 x - e^{7 x})

Simplify

3 \cdot \frac{1}{49} (e^{7 x} \cdot 7 x - e^{7 x}) : \frac{3}{49} (7 e^{7 x} x - e^{7 x})

= \frac{3}{49} (7 e^{7 x} x - e^{7 x})

Add a constant to the solution

= \frac{3}{49} (7 e^{7 x} x - e^{7 x}) + C

(y - (x^{2} + y^{2})) d x - x d y = 0

\frac{\partial}{\partial y} (ln (x + y + z))

\frac{5 e ^{- s}}{s + 1}

\int_{0}^{3} π (x)^{2} d x

\int_{1}^{3} 2 π x (\frac{e ^{x}}{x}) d x

What is the integral of 3xe^{7x} ?
The integral of 3xe^{7x} is 3/49 (7e^{7x}x-e^{7x})+C