What is the integral of ze^{2z} ?

The integral of ze^{2z} is 1/4 (e^{2z}*2z-e^{2z})+C

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integral of ze^{2z}

\int z e^{2 z} d z

\frac{1}{4} (e^{2 z} \cdot 2 z - e^{2 z}) + C

Solution steps

\int z e^{2 z} d z

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 2 z

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

Add a constant to the solution

= \frac{1}{4} (e^{2 z} \cdot 2 z - e^{2 z}) + C

y^{^{''}} = 6 t^{2}

\int_{0}^{6} x^{3} x^{2} + 36 d x

x \to - 3 - lim (\frac{( x + 2 )}{x + 3})

x (e^{\frac{y}{x}} - 1) y^{^{'}} = e^{\frac{y}{x}} (y - x), y (1) = 0

\frac{d}{d x} ((x^{2} + y^{2}) ln (x + y))

What is the integral of ze^{2z} ?
The integral of ze^{2z} is 1/4 (e^{2z}*2z-e^{2z})+C