What is the integral of (e^x+1)^3e^x ?

The integral of (e^x+1)^3e^x is ((e^x+1)^4)/4+C

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integral of (e^x+1)^3e^x

\int (e^{x} + 1)^{3} e^{x} d x

\frac{( e ^{x} + 1 ) ^{4}}{4} + C

Solution steps

\int (e^{x} + 1)^{3} e^{x} d x

Apply u-substitution

: \int u^{3} d u

= \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{u ^{4}}{4}

Substitute back

u = e^{x} + 1

= \frac{( e ^{x} + 1 ) ^{4}}{4}

Add a constant to the solution

= \frac{( e ^{x} + 1 ) ^{4}}{4} + C

\int e^{i 8 x} cos (5 x) d x

x \to 0 + lim (x cot (x) csc (x))

x^{4}, at x = 3

\frac{d}{d x} (\frac{16}{( 1 + 2 x ) ^{3}})

arctan (cos (3 x))

What is the integral of (e^x+1)^3e^x ?
The integral of (e^x+1)^3e^x is ((e^x+1)^4)/4+C