What is the integral of (5-e^{2x})^2 ?

The integral of (5-e^{2x})^2 is 1/2 (50x-10e^{2x}+1/2 e^{4x})+C

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integral of (5-e^{2x})^2

\int (5 - e^{2 x})^{2} d x

\frac{1}{2} (50 x - 10 e^{2 x} + \frac{1}{2} e^{4 x}) + C

Solution steps

\int (5 - e^{2 x})^{2} d x

Factor

5 - e^{2 x} : - (- 5 + e^{2 x})

= \int (- (- 5 + e^{2 x}))^{2} d x

Simplify

(- (- 5 + e^{2 x}))^{2} : (- 5 + e^{2 x})^{2}

= \int (- 5 + e^{2 x})^{2} d x

Apply u-substitution

= \int \frac{( - 5 + e ^{u} ) ^{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 (- 5 + e^{u})^{2} d u

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{( - 5 + v ) ^{2}}{v} d v

Expand

\frac{( - 5 + v ) ^{2}}{v} : \frac{25}{v} - 10 + v

= \frac{1}{2} \cdot \int \frac{25}{v} - 10 + 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 \frac{25}{v} d v - \int 10 d v + \int v d v)

\int \frac{25}{v} d v = 25 ln ∣ v ∣

\int 10 d v = 10 v

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

= \frac{1}{2} (25 ln ∣ v ∣ - 10 v + \frac{v ^{2}}{2})

Substitute back

= \frac{1}{2} (25 ln e^{2 x} - 10 e^{2 x} + \frac{( e ^{2 x} ) ^{2}}{2})

Simplify

\frac{1}{2} (25 ln e^{2 x} - 10 e^{2 x} + \frac{( e ^{2 x} ) ^{2}}{2}) : \frac{1}{2} (50 x - 10 e^{2 x} + \frac{1}{2} e^{4 x})

= \frac{1}{2} (50 x - 10 e^{2 x} + \frac{1}{2} e^{4 x})

Add a constant to the solution

= \frac{1}{2} (50 x - 10 e^{2 x} + \frac{1}{2} e^{4 x}) + C

What is the integral of (5-e^{2x})^2 ?
The integral of (5-e^{2x})^2 is 1/2 (50x-10e^{2x}+1/2 e^{4x})+C