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

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

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

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

e^{2} - 1

Decimal Notation

6.38905 \dots

Solution steps

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

Apply u-substitution

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

= \int_{0}^{2} e^{u} d u

Use the common integral:

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

= [e^{u}]_{0}^{2}

Compute the boundaries:

e^{2} - 1

= e^{2} - 1

\frac{\partial}{\partial x} (y^{2} arctan (x))

\frac{\partial}{\partial x} (\frac{x + y}{y - x})

x \to 0 lim (- x^{2} + 1)

\int_{1}^{25} 5 x e^{2 x} d x

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

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