What is the integral of ((sqrt(x^2-4)))/(x^6) ?

The integral of ((sqrt(x^2-4)))/(x^6) is -(sqrt(x^2-4))/(5x^5)+((x^2+2)sqrt(x^2-4))/(120x^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((sqrt(x^2-4)))/(x^6)

\int \frac{( x ^{2} - 4 )}{x ^{6}} d x

- \frac{x ^{2} - 4}{5 x ^{5}} + \frac{( x ^{2} + 2 ) x ^{2} - 4}{120 x ^{3}} + C

Solution steps

\int \frac{x ^{2} - 4}{x ^{6}} d x

Apply Integration By Parts

: - \frac{x ^{2} - 4}{5 x ^{5}} - \int - \frac{1}{5 x ^{4} x ^{2} - 4} d x

= - \frac{x ^{2} - 4}{5 x ^{5}} - \int - \frac{1}{5 x ^{4} x ^{2} - 4} d x

\int - \frac{1}{5 x ^{4} x ^{2} - 4} d x = - \frac{( x ^{2} + 2 ) x ^{2} - 4}{120 x ^{3}}

= - \frac{x ^{2} - 4}{5 x ^{5}} - (- \frac{( x ^{2} + 2 ) x ^{2} - 4}{120 x ^{3}})

Simplify

= - \frac{x ^{2} - 4}{5 x ^{5}} + \frac{( x ^{2} + 2 ) x ^{2} - 4}{120 x ^{3}}

Add a constant to the solution

= - \frac{x ^{2} - 4}{5 x ^{5}} + \frac{( x ^{2} + 2 ) x ^{2} - 4}{120 x ^{3}} + C

\frac{\partial}{\partial x} (5 x y + 2 x - x^{2} - 4 y^{2})

\int 8 tan^{7} (x) sec^{2} (x) d x

\frac{d y}{d x} = 8 - y

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

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

What is the integral of ((sqrt(x^2-4)))/(x^6) ?
The integral of ((sqrt(x^2-4)))/(x^6) is -(sqrt(x^2-4))/(5x^5)+((x^2+2)sqrt(x^2-4))/(120x^3)+C