What is the integral of 8x(4x^2+9)^4 ?

The integral of 8x(4x^2+9)^4 is 1/5 (4x^2+9)^5+C

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integral of 8x(4x^2+9)^4

\int 8 x (4 x^{2} + 9)^{4} d x

\frac{1}{5} (4 x^{2} + 9)^{5} + C

Solution steps

\int 8 x (4 x^{2} + 9)^{4} d x

Take the constant out:

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

= 8 \cdot \int x (4 x^{2} + 9)^{4} d x

Apply u-substitution

= 8 \cdot \int \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

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

Substitute back

u = 4 x^{2} + 9

= 8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 9 ) ^{5}}{5}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} + 9 ) ^{5}}{5} : \frac{1}{5} (4 x^{2} + 9)^{5}

= \frac{1}{5} (4 x^{2} + 9)^{5}

Add a constant to the solution

= \frac{1}{5} (4 x^{2} + 9)^{5} + C

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What is the integral of 8x(4x^2+9)^4 ?
The integral of 8x(4x^2+9)^4 is 1/5 (4x^2+9)^5+C