What is the integral of 2xe^{17x} ?

The integral of 2xe^{17x} is 2/289 (17e^{17x}x-e^{17x})+C

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integral of 2xe^{17x}

\int 2 x e^{17 x} d x

\frac{2}{289} (17 e^{17 x} x - e^{17 x}) + C

Solution steps

\int 2 x e^{17 x} d x

Take the constant out:

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

= 2 \cdot \int x e^{17 x} d x

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{289} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{289} (e^{u} u - \int e^{u} d u)

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

= 2 \cdot \frac{1}{289} (e^{u} u - e^{u})

Substitute back

u = 17 x

= 2 \cdot \frac{1}{289} (e^{17 x} \cdot 17 x - e^{17 x})

Simplify

2 \cdot \frac{1}{289} (e^{17 x} \cdot 17 x - e^{17 x}) : \frac{2}{289} (17 e^{17 x} x - e^{17 x})

= \frac{2}{289} (17 e^{17 x} x - e^{17 x})

Add a constant to the solution

= \frac{2}{289} (17 e^{17 x} x - e^{17 x}) + C

\frac{\partial}{\partial s} (r^{2} + s^{2})

n = 5 \sum \infty tan (\frac{1}{n})

\frac{d}{d x} (2 x - 2 sin (x))

f (x) = x^{4} + 8 x^{2} - x, (1, 8)

\frac{d}{d x} (y e^{x y})

What is the integral of 2xe^{17x} ?
The integral of 2xe^{17x} is 2/289 (17e^{17x}x-e^{17x})+C