What is the integral of (3-2x)^{100} ?

The integral of (3-2x)^{100} is 1/202 (-3+2x)^{101}+C

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integral of (3-2x)^{100}

\int (3 - 2 x)^{100} d x

\frac{1}{202} (- 3 + 2 x)^{101} + C

Solution steps

\int (3 - 2 x)^{100} d x

Factor

3 - 2 x : - (- 3 + 2 x)

= \int (- (- 3 + 2 x))^{100} d x

Simplify

(- (- 3 + 2 x))^{100} : (- 3 + 2 x)^{100}

= \int (- 3 + 2 x)^{100} d x

Apply u-substitution

: \int \frac{u ^{100}}{2} d u

= \int \frac{u ^{100}}{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^{100} d u

Apply the Power Rule

: \frac{u ^{101}}{101}

= \frac{1}{2} \cdot \frac{u ^{101}}{101}

Substitute back

u = - 3 + 2 x

= \frac{1}{2} \cdot \frac{( - 3 + 2 x ) ^{101}}{101}

Simplify

\frac{1}{2} \cdot \frac{( - 3 + 2 x ) ^{101}}{101} : \frac{1}{202} (- 3 + 2 x)^{101}

= \frac{1}{202} (- 3 + 2 x)^{101}

Add a constant to the solution

= \frac{1}{202} (- 3 + 2 x)^{101} + C

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What is the integral of (3-2x)^{100} ?
The integral of (3-2x)^{100} is 1/202 (-3+2x)^{101}+C