What is the integral of 1/(x^4sqrt(16-25x^2)) ?

The integral of 1/(x^4sqrt(16-25x^2)) is -((25x^2+8)sqrt(-25x^2+16))/(384x^3)+C

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integral of 1/(x^4sqrt(16-25x^2))

\int \frac{1}{x ^{4} 16 - 25 x ^{2}} d x

- \frac{( 25 x ^{2} + 8 ) - 25 x ^{2} + 16}{384 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} 16 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{125}{256 sin ^{4} ( u )} d u

= \int \frac{125}{256 sin ^{4} ( u )} d u

Take the constant out:

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

= \frac{125}{256} \cdot \int \frac{1}{sin ^{4} ( u )} d u

Rewrite using trig identities

= \frac{125}{256} \cdot \int csc^{4} (u) d u

Apply Integral Reduction

= \frac{125}{256} (- \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{125}{256} (- \frac{cos ( u ) csc ^{3} ( u )}{3} + \frac{2}{3} (- cot (u)))

Substitute back

u = arcsin (\frac{5}{4} x)

= \frac{125}{256} (- \frac{cos ( arcsin ( \frac{5}{4} x ) ) csc ^{3} ( arcsin ( \frac{5}{4} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{5}{4} x))))

Simplify

\frac{125}{256} (- \frac{cos ( arcsin ( \frac{5}{4} x ) ) csc ^{3} ( arcsin ( \frac{5}{4} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{5}{4} x)))) : - \frac{( 25 x ^{2} + 8 ) - 25 x ^{2} + 16}{384 x ^{3}}

= - \frac{( 25 x ^{2} + 8 ) - 25 x ^{2} + 16}{384 x ^{3}}

Add a constant to the solution

= - \frac{( 25 x ^{2} + 8 ) - 25 x ^{2} + 16}{384 x ^{3}} + C

\frac{d}{d x} (\frac{x ^{2} - 16}{x + 4})

(x^{2} + 3) (x^{2} - 3) x^{2}

\int sec^{2} (x) tan^{2} (x) d x

5000 (1 - \frac{1}{40} t)^{2}

\frac{\partial}{\partial x} (- \frac{1}{x - y})

What is the integral of 1/(x^4sqrt(16-25x^2)) ?
The integral of 1/(x^4sqrt(16-25x^2)) is -((25x^2+8)sqrt(-25x^2+16))/(384x^3)+C