What is the integral of 5xe^{6x} ?

The integral of 5xe^{6x} is 5/36 (6e^{6x}x-e^{6x})+C

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integral of 5xe^{6x}

\int 5 x e^{6 x} d x

\frac{5}{36} (6 e^{6 x} x - e^{6 x}) + C

Solution steps

\int 5 x e^{6 x} d x

Take the constant out:

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

= 5 \cdot \int x e^{6 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{36} d u

= 5 \cdot \int \frac{e ^{u} u}{36} d u

Take the constant out:

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

= 5 \cdot \frac{1}{36} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{36} (e^{u} u - \int e^{u} d u)

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

= 5 \cdot \frac{1}{36} (e^{u} u - e^{u})

Substitute back

u = 6 x

= 5 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x})

Simplify

5 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x}) : \frac{5}{36} (6 e^{6 x} x - e^{6 x})

= \frac{5}{36} (6 e^{6 x} x - e^{6 x})

Add a constant to the solution

= \frac{5}{36} (6 e^{6 x} x - e^{6 x}) + C

(e^{\frac{1}{x}} - \frac{e ^{\frac{1}{x}}}{x} - \frac{5 e ^{\frac{1}{x}}}{x ^{2}})^{^{'}}

t \to 0 lim ((t - 2)^{5})

\frac{( x - 1 )}{( x + 2 )}

y^{^{'}} = \frac{4 x ln ( 2 x )}{sin ( y )}

\frac{d}{d x} (2 e^{x} + 3 x)

What is the integral of 5xe^{6x} ?
The integral of 5xe^{6x} is 5/36 (6e^{6x}x-e^{6x})+C