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

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

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integral of e^{2x}*2x

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

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

Solution steps

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

Take the constant out:

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

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

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 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x + 3 ) ( x ^{2} - 3 x + 9 )}{x ^{3}})

\frac{d y}{d x} = 4 y^{2} - 3 y + 1

x = y + 12, x = y^{2} + 2 y

sin (tan (7 x))

\int_{2}^{4} (10 + 4 x - 3 x^{3}) d x

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