What is the integral of 3xe^{9x} ?

The integral of 3xe^{9x} is 1/27 (9e^{9x}x-e^{9x})+C

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integral of 3xe^{9x}

\int 3 x e^{9 x} d x

\frac{1}{27} (9 e^{9 x} x - e^{9 x}) + C

Solution steps

\int 3 x e^{9 x} d x

Take the constant out:

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

= 3 \cdot \int x e^{9 x} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{81} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{81} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{81} (e^{u} u - e^{u})

Substitute back

u = 9 x

= 3 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x})

Simplify

3 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x}) : \frac{1}{27} (9 e^{9 x} x - e^{9 x})

= \frac{1}{27} (9 e^{9 x} x - e^{9 x})

Add a constant to the solution

= \frac{1}{27} (9 e^{9 x} x - e^{9 x}) + C

\frac{d}{d x} (\frac{50}{x})

\frac{d y}{d x} - 2 x y = - 3 x^{2} e^{x^{2}}

y \to 5 lim (\frac{25 - y ^{2}}{y ^{2} - 4 y - 5})

\int \frac{sin ( 2 t + 1 )}{cos ^{2} ( 2 t + 1 )} d t

y^{^{'}} = \frac{1 - x}{y}

What is the integral of 3xe^{9x} ?
The integral of 3xe^{9x} is 1/27 (9e^{9x}x-e^{9x})+C