What is the integral from 0 to 1 of (x^2+2x+3)e^{2x} ?

The integral from 0 to 1 of (x^2+2x+3)e^{2x} is (9e^2-5)/4

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integral from 0 to 1 of (x^2+2x+3)e^{2x}

\int_{0}^{1} (x^{2} + 2 x + 3) e^{2 x} d x

\frac{9 e ^{2} - 5}{4}

Decimal Notation

15.37537 \dots

Solution steps

\int_{0}^{1} (x^{2} + 2 x + 3) e^{2 x} d x

Expand

(x^{2} + 2 x + 3) e^{2 x} : e^{2 x} x^{2} + 2 e^{2 x} x + 3 e^{2 x}

= \int_{0}^{1} e^{2 x} x^{2} + 2 e^{2 x} x + 3 e^{2 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

= \int_{0}^{1} e^{2 x} x^{2} d x + \int_{0}^{1} 2 e^{2 x} x d x + \int_{0}^{1} 3 e^{2 x} d x

\int_{0}^{1} e^{2 x} x^{2} d x = \frac{e ^{2} - 1}{4}

\int_{0}^{1} 2 e^{2 x} x d x = \frac{e ^{2} + 1}{2}

\int_{0}^{1} 3 e^{2 x} d x = \frac{3 ( e ^{2} - 1 )}{2}

= \frac{e ^{2} - 1}{4} + \frac{e ^{2} + 1}{2} + \frac{3 ( e ^{2} - 1 )}{2}

Simplify

\frac{e ^{2} - 1}{4} + \frac{e ^{2} + 1}{2} + \frac{3 ( e ^{2} - 1 )}{2} : \frac{9 e ^{2} - 5}{4}

= \frac{9 e ^{2} - 5}{4}

y = ln (x^{4} (x))

f (x) = 3 x - x^{2}

\frac{d}{d x} (\frac{1}{ln ( ln ( ln ( x ) ) )})

f (x) = (2 x^{- 2} + 3)^{- 3}

tan (2 t)

What is the integral from 0 to 1 of (x^2+2x+3)e^{2x} ?
The integral from 0 to 1 of (x^2+2x+3)e^{2x} is (9e^2-5)/4