What is the integral of 5xe^{2x} ?

The integral of 5xe^{2x} is 5/4 (2e^{2x}x-e^{2x})+C

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integral of 5xe^{2x}

\int 5 x e^{2 x} d x

\frac{5}{4} (2 e^{2 x} x - e^{2 x}) + C

Solution steps

\int 5 x e^{2 x} d x

Take the constant out:

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

= 5 \cdot \int x e^{2 x} d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= 5 \cdot \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = 2 x

= 5 \cdot \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

Simplify

5 \cdot \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) : \frac{5}{4} (2 e^{2 x} x - e^{2 x})

= \frac{5}{4} (2 e^{2 x} x - e^{2 x})

Add a constant to the solution

= \frac{5}{4} (2 e^{2 x} x - e^{2 x}) + C

arcsec (4 x)

y = - \frac{5}{( x + 3 ) ^{2}}

f (x) = 2 x + 7

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

\frac{d}{d x} (- k e^{l (x - m)} + n e^{- p (x - m)} + q)

What is the integral of 5xe^{2x} ?
The integral of 5xe^{2x} is 5/4 (2e^{2x}x-e^{2x})+C