What is the integral of (1-e^{3x})^2 ?

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

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integral of (1-e^{3x})^2

\int (1 - e^{3 x})^{2} d x

\frac{1}{6} (e^{6 x} - 4 e^{3 x} + 6 x) + C

Solution steps

\int (1 - e^{3 x})^{2} d x

Apply u-substitution

: \int - \frac{u ^{2}}{3 ( - u + 1 )} d u

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

\int 2 d v = 2 v

\int \frac{1}{v} d v = ln ∣ v ∣

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

Substitute back

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

Simplify

- \frac{1}{3} (- (\frac{( - ( 1 - e ^{3 x} ) + 1 ) ^{2}}{2} - 2 (- (1 - e^{3 x}) + 1) + ln - (1 - e^{3 x}) + 1)) : \frac{1}{6} (e^{6 x} - 4 e^{3 x} + 6 x)

= \frac{1}{6} (e^{6 x} - 4 e^{3 x} + 6 x)

Add a constant to the solution

= \frac{1}{6} (e^{6 x} - 4 e^{3 x} + 6 x) + C

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