What is the integral of 8xe^{9x} ?

The integral of 8xe^{9x} is 8/81 (9e^{9x}x-e^{9x})+C

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integral of 8xe^{9x}

\int 8 x e^{9 x} d x

\frac{8}{81} (9 e^{9 x} x - e^{9 x}) + C

Solution steps

\int 8 x e^{9 x} d x

Take the constant out:

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

= 8 \cdot \int x e^{9 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{81} d u

= 8 \cdot \int \frac{e ^{u} u}{81} d u

Take the constant out:

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

= 8 \cdot \frac{1}{81} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 8 \cdot \frac{1}{81} (e^{u} u - \int e^{u} d u)

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

= 8 \cdot \frac{1}{81} (e^{u} u - e^{u})

Substitute back

u = 9 x

= 8 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x})

Simplify

8 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x}) : \frac{8}{81} (9 e^{9 x} x - e^{9 x})

= \frac{8}{81} (9 e^{9 x} x - e^{9 x})

Add a constant to the solution

= \frac{8}{81} (9 e^{9 x} x - e^{9 x}) + C

x, 8

x \to 0 lim (\frac{( 484 + x ) ^{0.5} - 22}{x})

f (x) = x^{4} - 10 x^{2} + 9, at x = 2

x \to 2 lim (\frac{3}{( x - 2 ) ^{2}})

\int 2 x + 1 d x

What is the integral of 8xe^{9x} ?
The integral of 8xe^{9x} is 8/81 (9e^{9x}x-e^{9x})+C