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} 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{\partial}{\partial x} (y^{6} cos (3 x))

\int \frac{x ^{3}}{3 ( 1 + x ^{2} )} d x

f (x) = 3 x^{2} - 4 x, at x = - 1

\frac{d}{d y} (x - 2 y)

\frac{s + 2}{( s + 1 ) ^{2}}

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