What is the integral of 9xe^{6x} ?

The integral of 9xe^{6x} is 1/4 (6e^{6x}x-e^{6x})+C

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integral of 9xe^{6x}

\int 9 x e^{6 x} d x

\frac{1}{4} (6 e^{6 x} x - e^{6 x}) + C

Solution steps

\int 9 x e^{6 x} d x

Take the constant out:

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

= 9 \cdot \int x e^{6 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{36} d u

= 9 \cdot \int \frac{e ^{u} u}{36} d u

Take the constant out:

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

= 9 \cdot \frac{1}{36} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{36} (e^{u} u - \int e^{u} d u)

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

= 9 \cdot \frac{1}{36} (e^{u} u - e^{u})

Substitute back

u = 6 x

= 9 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x})

Simplify

9 \cdot \frac{1}{36} (e^{6 x} \cdot 6 x - e^{6 x}) : \frac{1}{4} (6 e^{6 x} x - e^{6 x})

= \frac{1}{4} (6 e^{6 x} x - e^{6 x})

Add a constant to the solution

= \frac{1}{4} (6 e^{6 x} x - e^{6 x}) + C

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

x \to \infty lim (e^{p x} (a cos (q (x) x) + b sin (q (x) x)))

\frac{d y}{d x} = 4 + e^{y - 4 x + 5}

\int x^{2} \cdot e^{3 x} d x

\frac{d}{d x} (4 3 x^{2} - 5 x + 2)

What is the integral of 9xe^{6x} ?
The integral of 9xe^{6x} is 1/4 (6e^{6x}x-e^{6x})+C