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

The integral of xe^{9x} is 1/81 (e^{9x}*9x-e^{9x})+C

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integral of xe^{9x}

\int x e^{9 x} d x

\frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x}) + C

Solution steps

\int x e^{9 x} d x

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 9 x

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

Add a constant to the solution

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

\int e^{\frac{z}{y}} d z

\frac{d}{d x} (\frac{3}{x ^{- 2}})

\frac{\partial}{\partial x} (1 + x^{2} y - x y)

\frac{d}{d t} (2 t + 2)

\frac{d}{d x} (3 (x + 1)^{2})

What is the integral of xe^{9x} ?
The integral of xe^{9x} is 1/81 (e^{9x}*9x-e^{9x})+C