What is the integral of (x^2)/((9x^2-4)^{5/2)} ?

The integral of (x^2)/((9x^2-4)^{5/2)} is -(x^3)/(12(9x^2-4)^{3/2)}+C

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integral of (x^2)/((9x^2-4)^{5/2)}

\int \frac{x ^{2}}{( 9 x ^{2} - 4 ) ^{\frac{5}{2}}} d x

- \frac{x ^{3}}{12 ( 9 x ^{2} - 4 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{x ^{2}}{( 9 x ^{2} - 4 ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{sec ^{3} ( u )}{108 tan ^{4} ( u )} d u

= \int \frac{sec ^{3} ( u )}{108 tan ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{108} \cdot \int \frac{sec ^{3} ( u )}{tan ^{4} ( u )} d u

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

= \frac{1}{108} \cdot \int cot^{4} (u) sec^{3} (u) d u

Rewrite using trig identities

= \frac{1}{108} \cdot \int \frac{cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{4}} d v

= \frac{1}{108} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

: - \frac{1}{3 v ^{3}}

= \frac{1}{108} (- \frac{1}{3 v ^{3}})

Substitute back

= \frac{1}{108} (- \frac{1}{3 sin ^{3} ( arcsec ( \frac{3}{2} x ) )})

Simplify

\frac{1}{108} (- \frac{1}{3 sin ^{3} ( arcsec ( \frac{3}{2} x ) )}) : - \frac{x ^{3}}{12 ( 9 x ^{2} - 4 ) ^{\frac{3}{2}}}

= - \frac{x ^{3}}{12 ( 9 x ^{2} - 4 ) ^{\frac{3}{2}}}

Add a constant to the solution

= - \frac{x ^{3}}{12 ( 9 x ^{2} - 4 ) ^{\frac{3}{2}}} + C

\int_{0}^{π} (5 e^{x} + 3 sin (x)) d x

\frac{\partial}{\partial x} (3 t sin^{2} (t))

f (x) = 5 x - 1

x^{4} y^{^{'}} + 3 x^{3} y = x^{2} e^{x}

\frac{d}{d x} (\frac{3 x - 9}{2 x + 4})

What is the integral of (x^2)/((9x^2-4)^{5/2)} ?
The integral of (x^2)/((9x^2-4)^{5/2)} is -(x^3)/(12(9x^2-4)^{3/2)}+C