What is the integral of (3x-1)^{2003} ?

The integral of (3x-1)^{2003} is 1/6012 (3x-1)^{2004}+C

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integral of (3x-1)^{2003}

\int (3 x - 1)^{2003} d x

\frac{1}{6012} (3 x - 1)^{2004} + C

Solution steps

\int (3 x - 1)^{2003} d x

Apply u-substitution

: \int \frac{u ^{2003}}{3} d u

= \int \frac{u ^{2003}}{3} 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 u^{2003} d u

Apply the Power Rule

: \frac{u ^{2004}}{2004}

= \frac{1}{3} \cdot \frac{u ^{2004}}{2004}

Substitute back

u = 3 x - 1

= \frac{1}{3} \cdot \frac{( 3 x - 1 ) ^{2004}}{2004}

Simplify

\frac{1}{3} \cdot \frac{( 3 x - 1 ) ^{2004}}{2004} : \frac{1}{6012} (3 x - 1)^{2004}

= \frac{1}{6012} (3 x - 1)^{2004}

Add a constant to the solution

= \frac{1}{6012} (3 x - 1)^{2004} + C

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What is the integral of (3x-1)^{2003} ?
The integral of (3x-1)^{2003} is 1/6012 (3x-1)^{2004}+C