What is the integral of (3-2x)^5 ?

The integral of (3-2x)^5 is -1/12 (3-2x)^6+C

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integral of (3-2x)^5

\int (3 - 2 x)^{5} d x

- \frac{1}{12} (3 - 2 x)^{6} + C

Solution steps

\int (3 - 2 x)^{5} d x

Apply u-substitution

: \int - \frac{u ^{5}}{2} d u

= \int - \frac{u ^{5}}{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 u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= - \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = 3 - 2 x

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{12} (3 - 2 x)^{6} + C

\frac{\partial}{\partial x} (e^{(x + y)} - 1)

- sin (2 (3)) 3

7 (2 z e^{z} + e^{z} z^{2})

f (x) = arcsin (1 - x^{2})

x^{^{''}} + 9 x = 27