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

The integral of 1/(x^4sqrt(4-x^2)) is -((x^2+2)sqrt(-x^2+4))/(24x^3)+C

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

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

- \frac{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}} + C

Solution steps

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

Apply Trigonometric Substitution

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

= \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

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

Rewrite using trig identities

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

Apply Integral Reduction

= \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)

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

Substitute back

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

= \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

\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{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}}

= - \frac{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}}

Add a constant to the solution

= - \frac{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}} + C

\int \frac{e ^{5 x}}{5} d x

\int_{0}^{1} \frac{2 x}{( x ^{2} + 3 ) ^{2}} d x

f (x) = \frac{- 9 x ^{2}}{2} + 7 x + 2

\frac{\partial}{\partial y} (arctan (x^{4} y^{3}))

5 (t - 1)

What is the integral of 1/(x^4sqrt(4-x^2)) ?
The integral of 1/(x^4sqrt(4-x^2)) is -((x^2+2)sqrt(-x^2+4))/(24x^3)+C