What is the integral of (16)/(x^4sqrt(4-x^2)) ?

The integral of (16)/(x^4sqrt(4-x^2)) is -(4sqrt(-x^2+4)}{3x^3}-2/3*\frac{sqrt(4-x^2))/x+C

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integral of (16)/(x^4sqrt(4-x^2))

\int \frac{16}{x ^{4} 4 - x ^{2}} d x

- \frac{4 - x ^{2} + 4}{3 x ^{3}} - \frac{2}{3} \cdot \frac{4 - x ^{2}}{x} + C

Solution steps

\int \frac{16}{x ^{4} 4 - x ^{2}} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{x ^{4} 4 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 sin ^{4} ( u )} d u

= 16 \cdot \int \frac{1}{16 sin ^{4} ( u )} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int \frac{1}{sin ^{4} ( u )} d u

Rewrite using trig identities

= 16 \cdot \frac{1}{16} \cdot \int csc^{4} (u) d u

Apply Integral Reduction

= 16 \cdot \frac{1}{16} (- \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)

= 16 \cdot \frac{1}{16} (- \frac{cos ( u ) csc ^{3} ( u )}{3} + \frac{2}{3} (- cot (u)))

Substitute back

u = arcsin (\frac{1}{2} x)

= 16 \cdot \frac{1}{16} (- \frac{cos ( arcsin ( \frac{1}{2} x ) ) csc ^{3} ( arcsin ( \frac{1}{2} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{1}{2} x))))

Simplify

16 \cdot \frac{1}{16} (- \frac{cos ( arcsin ( \frac{1}{2} x ) ) csc ^{3} ( arcsin ( \frac{1}{2} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{1}{2} x)))) : - \frac{4 - x ^{2} + 4}{3 x ^{3}} - \frac{2}{3} \cdot \frac{4 - x ^{2}}{x}

= - \frac{4 - x ^{2} + 4}{3 x ^{3}} - \frac{2}{3} \cdot \frac{4 - x ^{2}}{x}

Add a constant to the solution

= - \frac{4 - x ^{2} + 4}{3 x ^{3}} - \frac{2}{3} \cdot \frac{4 - x ^{2}}{x} + C

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What is the integral of (16)/(x^4sqrt(4-x^2)) ?
The integral of (16)/(x^4sqrt(4-x^2)) is -(4sqrt(-x^2+4)}{3x^3}-2/3*\frac{sqrt(4-x^2))/x+C