What is the integral of (4x-3)^9 ?

The integral of (4x-3)^9 is 1/40 (4x-3)^{10}+C

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integral of (4x-3)^9

\int (4 x - 3)^{9} d x

\frac{1}{40} (4 x - 3)^{10} + C

Solution steps

\int (4 x - 3)^{9} d x

Apply u-substitution

: \int \frac{u ^{9}}{4} d u

= \int \frac{u ^{9}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= \frac{1}{4} \cdot \frac{u ^{10}}{10}

Substitute back

u = 4 x - 3

= \frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{10}}{10}

Simplify

\frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{10}}{10} : \frac{1}{40} (4 x - 3)^{10}

= \frac{1}{40} (4 x - 3)^{10}

Add a constant to the solution

= \frac{1}{40} (4 x - 3)^{10} + C

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What is the integral of (4x-3)^9 ?
The integral of (4x-3)^9 is 1/40 (4x-3)^{10}+C