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

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

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integral of (e^x-1)^2

integral

(e^{x} - 1)^{2}

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \int u - 2 + \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 d u - \int 2 d u + \int \frac{1}{u} d u

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + ln ∣ u ∣

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

x + 1 x

\frac{x ^{2} - 81}{x ^{4}}

\frac{x ^{2} - 1}{x ^{2}}

(x - 1) e^{x}

f (x) = arctan (x^{\frac{1}{2}})

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