What is the integral of ((16-9x^2)^{3/2})/(x^6) ?

The integral of ((16-9x^2)^{3/2})/(x^6) is -((16-9x^2)^{5/2})/(80x^5)+C

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integral of ((16-9x^2)^{3/2})/(x^6)

\int \frac{( 16 - 9 x ^{2} ) ^{\frac{3}{2}}}{x ^{6}} d x

- \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}} + C

Solution steps

\int \frac{( 16 - 9 x ^{2} ) ^{\frac{3}{2}}}{x ^{6}} d x

Apply Trigonometric Substitution

: \int \frac{243 cos ^{4} ( u )}{16 sin ^{6} ( u )} d u

= \int \frac{243 cos ^{4} ( u )}{16 sin ^{6} ( u )} d u

Take the constant out:

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

= \frac{243}{16} \cdot \int \frac{cos ^{4} ( u )}{sin ^{6} ( u )} d u

Rewrite using trig identities

= \frac{243}{16} \cdot \int cot^{4} (u) csc^{2} (u) d u

Apply u-substitution

: \int - v^{4} d v

= \frac{243}{16} \cdot \int - v^{4} d v

Take the constant out:

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

= \frac{243}{16} (- \int v^{4} d v)

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{243}{16} (- \frac{v ^{5}}{5})

Substitute back

= \frac{243}{16} (- \frac{cot ^{5} ( arcsin ( \frac{3}{4} x ) )}{5})

Simplify

\frac{243}{16} (- \frac{cot ^{5} ( arcsin ( \frac{3}{4} x ) )}{5}) : - \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}}

= - \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}}

Add a constant to the solution

= - \frac{( 16 - 9 x ^{2} ) ^{\frac{5}{2}}}{80 x ^{5}} + C

\int \frac{16}{( x + 2 ) ^{2}} d x

y^{^{'}} = y + x y

n = 1 \sum \infty (\frac{( 3 n - 1 )}{( 3 n )})^{n^{2}}

tan (x) sin^{2} (y) d x + cos^{2} (x) cot (y) d y = 0

n = 0 \sum \infty 1 + \frac{1}{n ^{2}}

What is the integral of ((16-9x^2)^{3/2})/(x^6) ?
The integral of ((16-9x^2)^{3/2})/(x^6) is -((16-9x^2)^{5/2})/(80x^5)+C