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

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

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

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

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

Solution steps

\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

= - \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)

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2} e^{2} + 2^{y}) = 5

4 x^{4} - 4 x^{2}, x^{2}

\frac{d}{d x} (\frac{x - 1}{2})

\frac{d}{d x} (\frac{2 x - 3}{( x ^{2} + 4 ) ^{2}})

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

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