What is the integral of ye^{4y} ?

The integral of ye^{4y} is 1/16 (e^{4y}*4y-e^{4y})+C

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integral of ye^{4y}

\int y e^{4 y} d y

\frac{1}{16} (e^{4 y} \cdot 4 y - e^{4 y}) + C

Solution steps

\int y e^{4 y} d y

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 y

= \frac{1}{16} (e^{4 y} \cdot 4 y - e^{4 y})

Add a constant to the solution

= \frac{1}{16} (e^{4 y} \cdot 4 y - e^{4 y}) + C

\frac{4 s e ^{- 3 s}}{( s ^{2} + 4 )}

\frac{d y}{d x}, y = 2 x - 5

n = 1 \sum \infty arctan (4 n)

\frac{2 s ^{2} + 1}{s + 3}

\int_{- 3}^{3} ∣ 1 - x ∣ d x

What is the integral of ye^{4y} ?
The integral of ye^{4y} is 1/16 (e^{4y}*4y-e^{4y})+C