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

The integral of (e^x+1)e^x is ((e^x+1)^2)/2+C

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integral of (e^x+1)e^x

\int (e^{x} + 1) e^{x} d x

\frac{( e ^{x} + 1 ) ^{2}}{2} + C

Solution steps

\int (e^{x} + 1) e^{x} d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{u ^{2}}{2}

Substitute back

u = e^{x} + 1

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

Add a constant to the solution

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

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What is the integral of (e^x+1)e^x ?
The integral of (e^x+1)e^x is ((e^x+1)^2)/2+C