What is the integral of 1/(x^4sqrt(9-x^2)) ?

The integral of 1/(x^4sqrt(9-x^2)) is (-9sqrt(-x^2+9)-2x^2sqrt(9-x^2))/(243x^3)+C

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integral of 1/(x^4sqrt(9-x^2))

\int \frac{1}{x ^{4} 9 - x ^{2}} d x

\frac{- 9 - x ^{2} + 9 - 2 x ^{2} 9 - x ^{2}}{243 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} 9 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{81 sin ^{4} ( u )} d u

= \int \frac{1}{81 sin ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int \frac{1}{sin ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{81} \cdot \int csc^{4} (u) d u

Apply Integral Reduction

= \frac{1}{81} (- \frac{cos ( u ) csc ^{3} ( u )}{3} + \frac{2}{3} \cdot \int csc^{2} (u) d u)

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{81} (- \frac{cos ( u ) csc ^{3} ( u )}{3} + \frac{2}{3} (- cot (u)))

Substitute back

u = arcsin (\frac{1}{3} x)

= \frac{1}{81} (- \frac{cos ( arcsin ( \frac{1}{3} x ) ) csc ^{3} ( arcsin ( \frac{1}{3} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{1}{3} x))))

Simplify

\frac{1}{81} (- \frac{cos ( arcsin ( \frac{1}{3} x ) ) csc ^{3} ( arcsin ( \frac{1}{3} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{1}{3} x)))) : \frac{- 9 - x ^{2} + 9 - 2 x ^{2} 9 - x ^{2}}{243 x ^{3}}

= \frac{- 9 - x ^{2} + 9 - 2 x ^{2} 9 - x ^{2}}{243 x ^{3}}

Add a constant to the solution

= \frac{- 9 - x ^{2} + 9 - 2 x ^{2} 9 - x ^{2}}{243 x ^{3}} + C

n = 1 \sum \infty 1 - \frac{3}{7 ^{n}}

\frac{d y}{d t} = \frac{1}{2} t

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{- 3 x}

\int 50 sin (120 π) x d x

x \cdot d y = d x, y (- 1) = 3

What is the integral of 1/(x^4sqrt(9-x^2)) ?
The integral of 1/(x^4sqrt(9-x^2)) is (-9sqrt(-x^2+9)-2x^2sqrt(9-x^2))/(243x^3)+C