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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 2 tan^{2} (u) d u

= \int 2 tan^{2} (u) d u

Take the constant out:

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

= 2 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= 2 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x}, x = y^{4}

y^{^{'}} - \frac{6}{x} y = x^{9} y^{8}

\int - \frac{1}{csc ( y )} d y

\frac{d}{d x} (cos^{2} (ln (x + 1)) - e^{x^{2}})

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