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

The integral of (e^x+1)^2e^x is ((e^x+1)^3)/3+C

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integral of (e^x+1)^2e^x

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

\frac{( e ^{x} + 1 ) ^{3}}{3} + C

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = e^{x} + 1

= \frac{( e ^{x} + 1 ) ^{3}}{3}

Add a constant to the solution

= \frac{( e ^{x} + 1 ) ^{3}}{3} + C

4 x + 12, 8 - x^{2}, x = - 2, x = 6

\frac{d y}{d x} = x + 3 y

\int_{- \infty}^{16} r e^{\frac{r}{4}} d r

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

\frac{d}{d x} (100 x^{\frac{3}{2}})

What is the integral of (e^x+1)^2e^x ?
The integral of (e^x+1)^2e^x is ((e^x+1)^3)/3+C