What is the integral of (2x^4-3)^5(8x^3) ?

The integral of (2x^4-3)^5(8x^3) is 1/6 (2x^4-3)^6+C

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integral of (2x^4-3)^5(8x^3)

\int (2 x^{4} - 3)^{5} (8 x^{3}) d x

\frac{1}{6} (2 x^{4} - 3)^{6} + C

Solution steps

\int (2 x^{4} - 3)^{5} \cdot 8 x^{3} d x

Take the constant out:

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

= 8 \cdot \int (2 x^{4} - 3)^{5} x^{3} d x

Apply u-substitution

: \int \frac{u ^{5}}{8} d u

= 8 \cdot \int \frac{u ^{5}}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int u^{5} d u

Apply the Power Rule

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

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

Substitute back

u = 2 x^{4} - 3

= 8 \cdot \frac{1}{8} \cdot \frac{( 2 x ^{4} - 3 ) ^{6}}{6}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{( 2 x ^{4} - 3 ) ^{6}}{6} : \frac{1}{6} (2 x^{4} - 3)^{6}

= \frac{1}{6} (2 x^{4} - 3)^{6}

Add a constant to the solution

= \frac{1}{6} (2 x^{4} - 3)^{6} + C

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What is the integral of (2x^4-3)^5(8x^3) ?
The integral of (2x^4-3)^5(8x^3) is 1/6 (2x^4-3)^6+C