What is the integral of (20x^4+8)(2x^5+4x)^{15} ?

The integral of (20x^4+8)(2x^5+4x)^{15} is 1/8 (2x^5+4x)^{16}+C

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integral of (20x^4+8)(2x^5+4x)^{15}

\int (20 x^{4} + 8) (2 x^{5} + 4 x)^{15} d x

\frac{1}{8} (2 x^{5} + 4 x)^{16} + C

Solution steps

\int (20 x^{4} + 8) (2 x^{5} + 4 x)^{15} d x

Apply u-substitution

: \int 2 u^{15} d u

= \int 2 u^{15} d u

Take the constant out:

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

= 2 \cdot \int u^{15} d u

Apply the Power Rule

: \frac{u ^{16}}{16}

= 2 \cdot \frac{u ^{16}}{16}

Substitute back

u = 2 x^{5} + 4 x

= 2 \cdot \frac{( 2 x ^{5} + 4 x ) ^{16}}{16}

Simplify

2 \cdot \frac{( 2 x ^{5} + 4 x ) ^{16}}{16} : \frac{1}{8} (2 x^{5} + 4 x)^{16}

= \frac{1}{8} (2 x^{5} + 4 x)^{16}

Add a constant to the solution

= \frac{1}{8} (2 x^{5} + 4 x)^{16} + C

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What is the integral of (20x^4+8)(2x^5+4x)^{15} ?
The integral of (20x^4+8)(2x^5+4x)^{15} is 1/8 (2x^5+4x)^{16}+C