What is the integral of 9xe^{17x} ?

The integral of 9xe^{17x} is 9/289 (17e^{17x}x-e^{17x})+C

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integral of 9xe^{17x}

\int 9 x e^{17 x} d x

\frac{9}{289} (17 e^{17 x} x - e^{17 x}) + C

Solution steps

\int 9 x e^{17 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^{17 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{289} d u

= 9 \cdot \int \frac{e ^{u} u}{289} d u

Take the constant out:

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

= 9 \cdot \frac{1}{289} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 9 \cdot \frac{1}{289} (e^{u} u - \int e^{u} d u)

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

= 9 \cdot \frac{1}{289} (e^{u} u - e^{u})

Substitute back

u = 17 x

= 9 \cdot \frac{1}{289} (e^{17 x} \cdot 17 x - e^{17 x})

Simplify

9 \cdot \frac{1}{289} (e^{17 x} \cdot 17 x - e^{17 x}) : \frac{9}{289} (17 e^{17 x} x - e^{17 x})

= \frac{9}{289} (17 e^{17 x} x - e^{17 x})

Add a constant to the solution

= \frac{9}{289} (17 e^{17 x} x - e^{17 x}) + C

\int cot^{3} (2 x) d x

(x^{2} + 4)^{7}

x \to a lim (2 - x)

y^{^{''}} - 6 y^{^{'}} + 9 y = 24 - 9 t - (12 t e^{3 t} + 4 e^{3 t})

x \to 4 - lim (x)

What is the integral of 9xe^{17x} ?
The integral of 9xe^{17x} is 9/289 (17e^{17x}x-e^{17x})+C