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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= \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

\int 4 r - r^{2} sin (θ) d r

- 6 x^{2} y + y^{^{'}} = - 10 x^{2}

\frac{1}{s ( s ^{2} + 2 s + 2 )}

\frac{5}{s ^{2} - 12} - \frac{4 s}{s ^{2} + 8}

\frac{\partial}{\partial t} (s t^{2})

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