What is the integral of (e^x-3)^2 ?

The integral of (e^x-3)^2 is 1/2 e^{2x}-6e^x+9x+C

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integral of (e^x-3)^2

\int (e^{x} - 3)^{2} d x

\frac{1}{2} e^{2 x} - 6 e^{x} + 9 x + C

Solution steps

\int (e^{x} - 3)^{2} d x

Apply u-substitution

: \int \frac{( u - 3 ) ^{2}}{u} d u

= \int \frac{( u - 3 ) ^{2}}{u} d u

Expand

\frac{( u - 3 ) ^{2}}{u} : u - 6 + \frac{9}{u}

= \int u - 6 + \frac{9}{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 6 d u + \int \frac{9}{u} d u

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

\int 6 d u = 6 u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= \frac{u ^{2}}{2} - 6 u + 9 ln ∣ u ∣

Substitute back

u = e^{x}

= \frac{( e ^{x} ) ^{2}}{2} - 6 e^{x} + 9 ln ∣ e^{x} ∣

Simplify

\frac{( e ^{x} ) ^{2}}{2} - 6 e^{x} + 9 ln ∣ e^{x} ∣ : \frac{1}{2} e^{2 x} - 6 e^{x} + 9 x

= \frac{1}{2} e^{2 x} - 6 e^{x} + 9 x

Add a constant to the solution

= \frac{1}{2} e^{2 x} - 6 e^{x} + 9 x + C

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What is the integral of (e^x-3)^2 ?
The integral of (e^x-3)^2 is 1/2 e^{2x}-6e^x+9x+C