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

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

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

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

- \frac{( 1 - x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{cos ^{2} ( u )}{sin ^{4} ( u )} d u

= \int \frac{cos ^{2} ( u )}{sin ^{4} ( u )} d u

Rewrite using trig identities

= \int \frac{cot ^{2} ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int - v^{2} d v

= \int - v^{2} d v

Take the constant out:

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

= - \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= - \frac{v ^{3}}{3}

Substitute back

= - \frac{cot ^{3} ( arcsin ( x ) )}{3}

Simplify

- \frac{cot ^{3} ( arcsin ( x ) )}{3} : - \frac{( 1 - x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}}

= - \frac{( 1 - x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}}

Add a constant to the solution

= - \frac{( 1 - x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + C

(x, y) \to (0, 0) lim (x y^{2})

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

\frac{d}{d x} (4 x - 3)

(x + 2 y^{2}) \cdot d x + x y \cdot d y = 0

\int \frac{x}{7 x ^{2} - 3} d x

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