What is the integral of (2x-x^2+7)e^2x ?

The integral of (2x-x^2+7)e^2x is e^2((2x^3)/3-(x^4)/4+(7x^2)/2)+C

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integral of (2x-x^2+7)e^2x

integral

(2 x - x^{2} + 7) e^{2} x

e^{2} (\frac{2 x ^{3}}{3} - \frac{x ^{4}}{4} + \frac{7 x ^{2}}{2}) + C

Solution steps

\int (2 x - x^{2} + 7) e^{2} x d x

Take the constant out:

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

= e^{2} \cdot \int (2 x - x^{2} + 7) x d x

Expand

(2 x - x^{2} + 7) x : 2 x^{2} - x^{3} + 7 x

= e^{2} \cdot \int 2 x^{2} - x^{3} + 7 x d x

Apply the Sum Rule:

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

= e^{2} (\int 2 x^{2} d x - \int x^{3} d x + \int 7 x d x)

\int 2 x^{2} d x = \frac{2 x ^{3}}{3}

\int x^{3} d x = \frac{x ^{4}}{4}

\int 7 x d x = \frac{7 x ^{2}}{2}

= e^{2} (\frac{2 x ^{3}}{3} - \frac{x ^{4}}{4} + \frac{7 x ^{2}}{2})

Add a constant to the solution

= e^{2} (\frac{2 x ^{3}}{3} - \frac{x ^{4}}{4} + \frac{7 x ^{2}}{2}) + C

ln (x^{2} + 1) + 6

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

\frac{( 2 x + 1 )}{( x + 2 )}

\frac{1}{x ( x ^{8} - 1 )}

\frac{( 8 x + 2 )}{x}

What is the integral of (2x-x^2+7)e^2x ?
The integral of (2x-x^2+7)e^2x is e^2((2x^3)/3-(x^4)/4+(7x^2)/2)+C