What is the integral of 2ye^{2y} ?

The integral of 2ye^{2y} is e^{2y}y-1/2 e^{2y}+C

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integral of 2ye^{2y}

\int 2 y e^{2 y} d y

e^{2 y} y - \frac{1}{2} e^{2 y} + C

Solution steps

\int 2 y e^{2 y} d y

Take the constant out:

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

= 2 \cdot \int y e^{2 y} d y

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= 2 \cdot \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= 2 \cdot \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = 2 y

= 2 \cdot \frac{1}{4} (e^{2 y} \cdot 2 y - e^{2 y})

Simplify

2 \cdot \frac{1}{4} (e^{2 y} \cdot 2 y - e^{2 y}) : e^{2 y} y - \frac{1}{2} e^{2 y}

= e^{2 y} y - \frac{1}{2} e^{2 y}

Add a constant to the solution

= e^{2 y} y - \frac{1}{2} e^{2 y} + C

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What is the integral of 2ye^{2y} ?
The integral of 2ye^{2y} is e^{2y}y-1/2 e^{2y}+C