What is the integral of e^{2x}(1-6e^{2x})^2 ?

The integral of e^{2x}(1-6e^{2x})^2 is -1/36 (1-6e^{2x})^3+C

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integral of e^{2x}(1-6e^{2x})^2

\int e^{2 x} (1 - 6 e^{2 x})^{2} d x

- \frac{1}{36} (1 - 6 e^{2 x})^{3} + C

Solution steps

\int e^{2 x} (1 - 6 e^{2 x})^{2} d x

Apply u-substitution

: \int - \frac{u ^{2}}{12} d u

= \int - \frac{u ^{2}}{12} d u

Take the constant out:

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

= - \frac{1}{12} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - \frac{1}{12} \cdot \frac{u ^{3}}{3}

Substitute back

u = 1 - 6 e^{2 x}

= - \frac{1}{12} \cdot \frac{( 1 - 6 e ^{2 x} ) ^{3}}{3}

Simplify

- \frac{1}{12} \cdot \frac{( 1 - 6 e ^{2 x} ) ^{3}}{3} : - \frac{1}{36} (1 - 6 e^{2 x})^{3}

= - \frac{1}{36} (1 - 6 e^{2 x})^{3}

Add a constant to the solution

= - \frac{1}{36} (1 - 6 e^{2 x})^{3} + C

\int \frac{1}{y ^{2} + 4} d y

z

x (x^{2} + 6) (x + 9)

f (x) = e^{4 x} cos (π x), at x = 0

cos (3 x)

What is the integral of e^{2x}(1-6e^{2x})^2 ?
The integral of e^{2x}(1-6e^{2x})^2 is -1/36 (1-6e^{2x})^3+C