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

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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

y = x^{3} - 3 x^{2} + 25

\int_{0}^{π} 9 e^{x} sin (x) d x

\frac{d}{d x} (arctan (sec (x) + tan (x)))

y = \frac{sin ( x )}{x ^{2}}

\frac{d}{d x} (\frac{5}{x + 1})

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