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

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

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integral of e^{2x}(x^2+2x)

\int e^{2 x} (x^{2} + 2 x) d x

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

Solution steps

\int e^{2 x} (x^{2} + 2 x) d x

Apply u-substitution

: \int \frac{1}{2} e^{u} (\frac{u ^{2}}{4} + u) d u

= \int \frac{1}{2} e^{u} (\frac{u ^{2}}{4} + u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{u} (\frac{u ^{2}}{4} + u) d u

Expand

e^{u} (\frac{u ^{2}}{4} + u) : \frac{e ^{u} u ^{2}}{4} + e^{u} u

= \frac{1}{2} \cdot \int \frac{e ^{u} u ^{2}}{4} + e^{u} u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{e ^{u} u ^{2}}{4} d u + \int e^{u} u d u)

\int \frac{e ^{u} u ^{2}}{4} d u = \frac{1}{4} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

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

= \frac{1}{2} (\frac{1}{4} (u^{2} e^{u} - 2 (e^{u} u - e^{u})) + e^{u} u - e^{u})

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

(t + 4)^{(\frac{2}{3})} (2 t^{2} - 4)^{3}

x \to 2 lim (x^{2} + 12)

\frac{d ^{2}}{d x ^{2}} (\frac{7}{8 x ^{6}})

f (x) = 6 x - 32 x, at x = 4

f (x) = \frac{x ^{2} + sec ( x )}{x - 2 e ^{x}}

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