What is the integral of x^3sqrt(4x^2-9) ?

The integral of x^3sqrt(4x^2-9) is 1/40 (2x^2+3)(4x^2-9)^{3/2}+C

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integral of x^3sqrt(4x^2-9)

\int x^{3} 4 x^{2} - 9 d x

\frac{1}{40} (2 x^{2} + 3) (4 x^{2} - 9)^{\frac{3}{2}} + C

Solution steps

\int x^{3} 4 x^{2} - 9 d x

Apply u-substitution

: \int \frac{u ( u + 9 )}{32} d u

= \int \frac{u ( u + 9 )}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int u (u + 9) d u

Expand

u (u + 9) : u^{\frac{3}{2}} + 9 u

= \frac{1}{32} \cdot \int u^{\frac{3}{2}} + 9 u 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

= \frac{1}{32} (\int u^{\frac{3}{2}} d u + \int 9 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{1}{32} (\frac{2}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}})

Substitute back

u = 4 x^{2} - 9

= \frac{1}{32} (\frac{2}{5} (4 x^{2} - 9)^{\frac{5}{2}} + 6 (4 x^{2} - 9)^{\frac{3}{2}})

Simplify

\frac{1}{32} (\frac{2}{5} (4 x^{2} - 9)^{\frac{5}{2}} + 6 (4 x^{2} - 9)^{\frac{3}{2}}) : \frac{1}{40} (2 x^{2} + 3) (4 x^{2} - 9)^{\frac{3}{2}}

= \frac{1}{40} (2 x^{2} + 3) (4 x^{2} - 9)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{40} (2 x^{2} + 3) (4 x^{2} - 9)^{\frac{3}{2}} + C

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What is the integral of x^3sqrt(4x^2-9) ?
The integral of x^3sqrt(4x^2-9) is 1/40 (2x^2+3)(4x^2-9)^{3/2}+C