What is the integral of 4xe^{6x} ?

The integral of 4xe^{6x} is 1/9 (6e^{6x}x-e^{6x})+C

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integral of 4xe^{6x}

\int 4 x e^{6 x} d x

\frac{1}{9} (6 e^{6 x} x - e^{6 x}) + C

Solution steps

\int 4 x e^{6 x} d x

Take the constant out:

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

= 4 \cdot \int x e^{6 x} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{36} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot \frac{1}{36} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{1}{36} (e^{u} u - e^{u})

Substitute back

u = 6 x

= 4 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x})

Simplify

4 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x}) : \frac{1}{9} (6 e^{6 x} x - e^{6 x})

= \frac{1}{9} (6 e^{6 x} x - e^{6 x})

Add a constant to the solution

= \frac{1}{9} (6 e^{6 x} x - e^{6 x}) + C

- \frac{1}{2} x^{2} + \frac{5}{2} x - 2, [- 2, 7]

f (x) = \frac{1 + x}{( 2 x ^{2} + 1 ) ^{2}}

y^{^{'}} + 2 y = 0, y (1) = 1

y y^{^{'}} = \frac{x}{x ^{2} + 1}

y^{^{'}} + y = e^{x}

What is the integral of 4xe^{6x} ?
The integral of 4xe^{6x} is 1/9 (6e^{6x}x-e^{6x})+C