What is the integral of 7xe^{6x} ?

The integral of 7xe^{6x} is 7/36 (6e^{6x}x-e^{6x})+C

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integral of 7xe^{6x}

\int 7 x e^{6 x} d x

\frac{7}{36} (6 e^{6 x} x - e^{6 x}) + C

Solution steps

\int 7 x e^{6 x} d x

Take the constant out:

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

= 7 \cdot \int x e^{6 x} d x

Apply u-substitution

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

= 7 \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

= 7 \cdot \frac{1}{36} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 7 \cdot \frac{1}{36} (e^{u} u - \int e^{u} d u)

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

= 7 \cdot \frac{1}{36} (e^{u} u - e^{u})

Substitute back

u = 6 x

= 7 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x})

Simplify

7 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x}) : \frac{7}{36} (6 e^{6 x} x - e^{6 x})

= \frac{7}{36} (6 e^{6 x} x - e^{6 x})

Add a constant to the solution

= \frac{7}{36} (6 e^{6 x} x - e^{6 x}) + C

y = 2 x^{2} + 1, (3, 19)

f (x) = (1 + 2 x)^{2}, at x = 4

\frac{x ^{3}}{2 - x ^{2}}

(sin^{4} (x))^{^{'}}

\frac{2}{( s + 1 ) ^{3}}

What is the integral of 7xe^{6x} ?
The integral of 7xe^{6x} is 7/36 (6e^{6x}x-e^{6x})+C