What is the integral of (e^{2x}*,x) ?

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

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integral of (e^{2x}*,x)

\int (e^{2 x} \cdot, x) d x

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

Solution steps

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

Take the constant out:

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

=, \int e^{2 x} x d x

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 x

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

Add a constant to the solution

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

\frac{8 \cdot x}{x ^{2} + p}

\int e^{x} \cdot e^{x} d x

n = 1 \sum \infty \frac{4}{5 ^{n}}

6 x + \frac{8}{3}

\int \frac{ln ^{2} ( 5 )}{2 x} d x

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