What is the integral of (16x^3+6)(2x^4+3x)^{19} ?

The integral of (16x^3+6)(2x^4+3x)^{19} is 1/10 (2x^4+3x)^{20}+C

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integral of (16x^3+6)(2x^4+3x)^{19}

\int (16 x^{3} + 6) (2 x^{4} + 3 x)^{19} d x

\frac{1}{10} (2 x^{4} + 3 x)^{20} + C

Solution steps

\int (16 x^{3} + 6) (2 x^{4} + 3 x)^{19} d x

Apply u-substitution

: \int 2 u^{19} d u

= \int 2 u^{19} d u

Take the constant out:

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

= 2 \cdot \int u^{19} d u

Apply the Power Rule

: \frac{u ^{20}}{20}

= 2 \cdot \frac{u ^{20}}{20}

Substitute back

u = 2 x^{4} + 3 x

= 2 \cdot \frac{( 2 x ^{4} + 3 x ) ^{20}}{20}

Simplify

2 \cdot \frac{( 2 x ^{4} + 3 x ) ^{20}}{20} : \frac{1}{10} (2 x^{4} + 3 x)^{20}

= \frac{1}{10} (2 x^{4} + 3 x)^{20}

Add a constant to the solution

= \frac{1}{10} (2 x^{4} + 3 x)^{20} + C

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What is the integral of (16x^3+6)(2x^4+3x)^{19} ?
The integral of (16x^3+6)(2x^4+3x)^{19} is 1/10 (2x^4+3x)^{20}+C