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

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

Popular >

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

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

\frac{e ^{3} - 1}{2}

Decimal Notation

9.54276 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{3} \frac{e ^{u}}{2} d u

= \int_{0}^{3} \frac{e ^{u}}{2} 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_{0}^{3} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{0}^{3}

Compute the boundaries:

e^{3} - 1

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

Simplify

= \frac{e ^{3} - 1}{2}

\int e^{t a n (5 x)} sec^{2} (5 x) d x

\frac{d y}{d x} = - \frac{( x + y ) ^{2}}{2 x y + x ^{2} - 1}

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

\frac{\partial}{\partial x} (8 x^{5} y^{6} + 6 x^{4} y^{8})

6 x^{- 4}

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