What is the integral of 6(4x-3)^7 ?

The integral of 6(4x-3)^7 is 3/16 (4x-3)^8+C

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integral of 6(4x-3)^7

\int 6 (4 x - 3)^{7} d x

\frac{3}{16} (4 x - 3)^{8} + C

Solution steps

\int 6 (4 x - 3)^{7} d x

Take the constant out:

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

= 6 \cdot \int (4 x - 3)^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{4} d u

= 6 \cdot \int \frac{u ^{7}}{4} d u

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= 6 \cdot \frac{1}{4} \cdot \frac{u ^{8}}{8}

Substitute back

u = 4 x - 3

= 6 \cdot \frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{8}}{8}

Simplify

6 \cdot \frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{8}}{8} : \frac{3}{16} (4 x - 3)^{8}

= \frac{3}{16} (4 x - 3)^{8}

Add a constant to the solution

= \frac{3}{16} (4 x - 3)^{8} + C

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What is the integral of 6(4x-3)^7 ?
The integral of 6(4x-3)^7 is 3/16 (4x-3)^8+C