What is the integral of 4xe^{9x} ?

The integral of 4xe^{9x} is 4/81 (9e^{9x}x-e^{9x})+C

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integral of 4xe^{9x}

\int 4 x e^{9 x} d x

\frac{4}{81} (9 e^{9 x} x - e^{9 x}) + C

Solution steps

\int 4 x e^{9 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^{9 x} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{81} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot \frac{1}{81} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{1}{81} (e^{u} u - e^{u})

Substitute back

u = 9 x

= 4 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x})

Simplify

4 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x}) : \frac{4}{81} (9 e^{9 x} x - e^{9 x})

= \frac{4}{81} (9 e^{9 x} x - e^{9 x})

Add a constant to the solution

= \frac{4}{81} (9 e^{9 x} x - e^{9 x}) + C

f (x) = x^{4} - 5 x^{2} - x, at x = 1

y = \frac{3}{x}, (5, \frac{3}{5})

f (x) = (2 x + 3) (x^{2} + 2 x)

\int_{0}^{5} x e^{- 0.6 x} d x

\frac{d}{d x} (cot (x))

What is the integral of 4xe^{9x} ?
The integral of 4xe^{9x} is 4/81 (9e^{9x}x-e^{9x})+C