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

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

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integral of ((1-x^2)^{1/2})/(x^4)

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

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

Solution steps

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

Apply Trigonometric Substitution

= \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

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}

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

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