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

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

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integral of (e^x+1)^3

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

\frac{1}{3} e^{3 x} + \frac{3}{2} e^{2 x} + 3 e^{x} + x + C

Solution steps

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

Apply u-substitution

: \int \frac{( u + 1 ) ^{3}}{u} d u

= \int \frac{( u + 1 ) ^{3}}{u} d u

Expand

\frac{( u + 1 ) ^{3}}{u} : u^{2} + 3 u + 3 + \frac{1}{u}

= \int u^{2} + 3 u + 3 + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{2} d u + \int 3 u d u + \int 3 d u + \int \frac{1}{u} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u d u = \frac{3 u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{3}}{3} + \frac{3 u ^{2}}{2} + 3 u + ln ∣ u ∣

Substitute back

u = e^{x}

= \frac{( e ^{x} ) ^{3}}{3} + \frac{3 ( e ^{x} ) ^{2}}{2} + 3 e^{x} + ln ∣ e^{x} ∣

Simplify

\frac{( e ^{x} ) ^{3}}{3} + \frac{3 ( e ^{x} ) ^{2}}{2} + 3 e^{x} + ln ∣ e^{x} ∣ : \frac{1}{3} e^{3 x} + \frac{3}{2} e^{2 x} + 3 e^{x} + x

= \frac{1}{3} e^{3 x} + \frac{3}{2} e^{2 x} + 3 e^{x} + x

Add a constant to the solution

= \frac{1}{3} e^{3 x} + \frac{3}{2} e^{2 x} + 3 e^{x} + x + C

\int (3 x^{2} - 6 x + \frac{9}{x} + 4 e^{x}) d x

y^{^{'}} = 0.05 y - 12 \cdot 4000

\frac{d}{d x} (x sin (x + y) + y cos (x))

\frac{d}{d t} (\frac{t}{1 + t ^{2}})

x^{2}, 1, 0

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