What is the integral of (e^{3x}-4)^2 ?

The integral of (e^{3x}-4)^2 is 1/3 (1/2 e^{6x}-8e^{3x}+48x)+C

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integral of (e^{3x}-4)^2

\int (e^{3 x} - 4)^{2} d x

\frac{1}{3} (\frac{1}{2} e^{6 x} - 8 e^{3 x} + 48 x) + C

Solution steps

\int (e^{3 x} - 4)^{2} d x

Apply u-substitution

: \int \frac{( e ^{u} - 4 ) ^{2}}{3} d u

= \int \frac{( e ^{u} - 4 ) ^{2}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (e^{u} - 4)^{2} d u

Apply u-substitution

: \int \frac{( v - 4 ) ^{2}}{v} d v

= \frac{1}{3} \cdot \int \frac{( v - 4 ) ^{2}}{v} d v

Expand

\frac{( v - 4 ) ^{2}}{v} : v - 8 + \frac{16}{v}

= \frac{1}{3} \cdot \int v - 8 + \frac{16}{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}{3} (\int v d v - \int 8 d v + \int \frac{16}{v} d v)

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

\int 8 d v = 8 v

\int \frac{16}{v} d v = 16 ln ∣ v ∣

= \frac{1}{3} (\frac{v ^{2}}{2} - 8 v + 16 ln ∣ v ∣)

Substitute back

= \frac{1}{3} (\frac{( e ^{3 x} ) ^{2}}{2} - 8 e^{3 x} + 16 ln e^{3 x})

Simplify

\frac{1}{3} (\frac{( e ^{3 x} ) ^{2}}{2} - 8 e^{3 x} + 16 ln e^{3 x}) : \frac{1}{3} (\frac{1}{2} e^{6 x} - 8 e^{3 x} + 48 x)

= \frac{1}{3} (\frac{1}{2} e^{6 x} - 8 e^{3 x} + 48 x)

Add a constant to the solution

= \frac{1}{3} (\frac{1}{2} e^{6 x} - 8 e^{3 x} + 48 x) + C

\frac{x}{x ^{2} + 64}

y = 3 x^{3} - x^{2} + 9, (2, 29)

\int (x - 1) sin (π x) d x

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What is the integral of (e^{3x}-4)^2 ?
The integral of (e^{3x}-4)^2 is 1/3 (1/2 e^{6x}-8e^{3x}+48x)+C