What is the integral of 17xe^{19x} ?

The integral of 17xe^{19x} is 17/361 (19e^{19x}x-e^{19x})+C

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integral of 17xe^{19x}

\int 17 x e^{19 x} d x

\frac{17}{361} (19 e^{19 x} x - e^{19 x}) + C

Solution steps

\int 17 x e^{19 x} d x

Take the constant out:

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

= 17 \cdot \int x e^{19 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{361} d u

= 17 \cdot \int \frac{e ^{u} u}{361} d u

Take the constant out:

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

= 17 \cdot \frac{1}{361} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 17 \cdot \frac{1}{361} (e^{u} u - \int e^{u} d u)

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

= 17 \cdot \frac{1}{361} (e^{u} u - e^{u})

Substitute back

u = 19 x

= 17 \cdot \frac{1}{361} (e^{19 x} \cdot 19 x - e^{19 x})

Simplify

17 \cdot \frac{1}{361} (e^{19 x} \cdot 19 x - e^{19 x}) : \frac{17}{361} (19 e^{19 x} x - e^{19 x})

= \frac{17}{361} (19 e^{19 x} x - e^{19 x})

Add a constant to the solution

= \frac{17}{361} (19 e^{19 x} x - e^{19 x}) + C

x \to 0 + lim ((tan (5 x))^{x})

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

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

\frac{d}{d x} (\frac{5}{x ^{9}})

\int_{- 4}^{4} 16 - x^{2} d x

What is the integral of 17xe^{19x} ?
The integral of 17xe^{19x} is 17/361 (19e^{19x}x-e^{19x})+C