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

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

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

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

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

Decimal

4.91280 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{e + 1} u d u

= \int_{2}^{e + 1} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{2}^{e + 1}

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

Compute the boundaries:

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

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

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