What is the integral of (sqrt(4x^2-16))/x ?

The integral of (sqrt(4x^2-16))/x is 4(-arcsec(x/2)+1/2 sqrt(x^2-4))+C

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integral of (sqrt(4x^2-16))/x

\int \frac{4 x ^{2} - 16}{x} d x

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

Solution steps

\int \frac{4 x ^{2} - 16}{x} d x

Apply Trigonometric Substitution

: \int 4 tan^{2} (u) d u

= \int 4 tan^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 4 (- u + tan (u))

Substitute back

u = arcsec (\frac{2}{4} x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((tan (x))^{- 1})

\int sec^{2} (x) tan^{4} (x) d x

x \to \infty lim (e^{\frac{- x}{100}})

x \to - 2 lim (\frac{x + 3}{x + 2})

(2 x^{2} + 2)^{3} (x^{2} - 1)^{2}

What is the integral of (sqrt(4x^2-16))/x ?
The integral of (sqrt(4x^2-16))/x is 4(-arcsec(x/2)+1/2 sqrt(x^2-4))+C