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

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

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

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

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

Solution steps

\int \frac{4 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

= 4 \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

= 4 (- \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)

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

Simplify

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

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

Add a constant to the solution

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

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