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

The integral of (e^{2x}-3)^2 is 1/2 (1/2 e^{4x}-6e^{2x}+18x)+C

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integral of (e^{2x}-3)^2

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

\frac{1}{2} (\frac{1}{2} e^{4 x} - 6 e^{2 x} + 18 x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int (e^{u} - 3)^{2} d u

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{( v - 3 ) ^{2}}{v} d v

Expand

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

= \frac{1}{2} \cdot \int v - 6 + \frac{9}{v} d v

Apply the Sum Rule:

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

= \frac{1}{2} (\int v d v - \int 6 d v + \int \frac{9}{v} d v)

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

\int 6 d v = 6 v

\int \frac{9}{v} d v = 9 ln ∣ v ∣

= \frac{1}{2} (\frac{v ^{2}}{2} - 6 v + 9 ln ∣ v ∣)

Substitute back

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

Simplify

\frac{1}{2} (\frac{( e ^{2 x} ) ^{2}}{2} - 6 e^{2 x} + 9 ln e^{2 x}) : \frac{1}{2} (\frac{1}{2} e^{4 x} - 6 e^{2 x} + 18 x)

= \frac{1}{2} (\frac{1}{2} e^{4 x} - 6 e^{2 x} + 18 x)

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} e^{4 x} - 6 e^{2 x} + 18 x) + C

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What is the integral of (e^{2x}-3)^2 ?
The integral of (e^{2x}-3)^2 is 1/2 (1/2 e^{4x}-6e^{2x}+18x)+C