What is the integral of (8x^3)/(sqrt(4x^2+9)) ?

The integral of (8x^3)/(sqrt(4x^2+9)) is 1/3 (2x^2-9)sqrt(4x^2+9)+C

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integral of (8x^3)/(sqrt(4x^2+9))

\int \frac{8 x ^{3}}{4 x ^{2} + 9} d x

\frac{1}{3} (2 x^{2} - 9) 4 x^{2} + 9 + C

Solution steps

\int \frac{8 x ^{3}}{4 x ^{2} + 9} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{3}}{4 x ^{2} + 9} d x

Apply u-substitution

: \int \frac{u ^{2} - 9}{16} d u

= 8 \cdot \int \frac{u ^{2} - 9}{16} d u

Take the constant out:

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

= 8 \cdot \frac{1}{16} \cdot \int u^{2} - 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 \cdot \frac{1}{16} (\int u^{2} d u - \int 9 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 9 d u = 9 u

= 8 \cdot \frac{1}{16} (\frac{u ^{3}}{3} - 9 u)

Substitute back

u = 4 x^{2} + 9

= 8 \cdot \frac{1}{16} (\frac{( 4 x ^{2} + 9 ) ^{3}}{3} - 9 4 x^{2} + 9)

Simplify

8 \cdot \frac{1}{16} (\frac{( 4 x ^{2} + 9 ) ^{3}}{3} - 9 4 x^{2} + 9) : \frac{1}{3} (2 x^{2} - 9) 4 x^{2} + 9

= \frac{1}{3} (2 x^{2} - 9) 4 x^{2} + 9

Add a constant to the solution

= \frac{1}{3} (2 x^{2} - 9) 4 x^{2} + 9 + C

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What is the integral of (8x^3)/(sqrt(4x^2+9)) ?
The integral of (8x^3)/(sqrt(4x^2+9)) is 1/3 (2x^2-9)sqrt(4x^2+9)+C