What is the integral of 7xe^{8x} ?

The integral of 7xe^{8x} is 7/64 (8e^{8x}x-e^{8x})+C

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integral of 7xe^{8x}

\int 7 x e^{8 x} d x

\frac{7}{64} (8 e^{8 x} x - e^{8 x}) + C

Solution steps

\int 7 x e^{8 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^{8 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{64} d u

= 7 \cdot \int \frac{e ^{u} u}{64} d u

Take the constant out:

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

= 7 \cdot \frac{1}{64} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 7 \cdot \frac{1}{64} (e^{u} u - \int e^{u} d u)

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

= 7 \cdot \frac{1}{64} (e^{u} u - e^{u})

Substitute back

u = 8 x

= 7 \cdot \frac{1}{64} (e^{8 x} \cdot 8 x - e^{8 x})

Simplify

7 \cdot \frac{1}{64} (e^{8 x} \cdot 8 x - e^{8 x}) : \frac{7}{64} (8 e^{8 x} x - e^{8 x})

= \frac{7}{64} (8 e^{8 x} x - e^{8 x})

Add a constant to the solution

= \frac{7}{64} (8 e^{8 x} x - e^{8 x}) + C

\frac{d}{d x} ((\frac{x + 6}{x - 6})^{5})

\int x (1 - x^{2})^{5} d x

x = - 5, x = 3, y = 2 x^{2} + 12, y = 0

\int \frac{1}{( x - 2 ) ( x - 5 )} d x

\int 6 x (4 x^{2} - 9)^{2} d x

What is the integral of 7xe^{8x} ?
The integral of 7xe^{8x} is 7/64 (8e^{8x}x-e^{8x})+C