What is the integral of (2sqrt(x^2-4))/(x^3) ?

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

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integral of (2sqrt(x^2-4))/(x^3)

\int \frac{2 x ^{2} - 4}{x ^{3}} d x

- \frac{x ^{2} - 4}{x ^{2}} + \frac{1}{2} arcsec (\frac{1}{2} x) + C

Solution steps

\int \frac{2 x ^{2} - 4}{x ^{3}} d x

Take the constant out:

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

= 2 \cdot \int \frac{x ^{2} - 4}{x ^{3}} d x

Apply Integration By Parts

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

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

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

= 2 (- \frac{x ^{2} - 4}{2 x ^{2}} - (- \frac{1}{4} arcsec (\frac{1}{2} x)))

Simplify

2 (- \frac{x ^{2} - 4}{2 x ^{2}} - (- \frac{1}{4} arcsec (\frac{1}{2} x))) : - \frac{x ^{2} - 4}{x ^{2}} + \frac{1}{2} arcsec (\frac{1}{2} x)

= - \frac{x ^{2} - 4}{x ^{2}} + \frac{1}{2} arcsec (\frac{1}{2} x)

Add a constant to the solution

= - \frac{x ^{2} - 4}{x ^{2}} + \frac{1}{2} arcsec (\frac{1}{2} x) + C

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