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

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

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integral of 1/(sqrt(16x^2-4))

\int \frac{1}{16 x ^{2} - 4} d x

\frac{1}{4} ln 4 x^{2} - 1 + 2 x + C

Solution steps

\int \frac{1}{16 x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4} d u

= \int \frac{sec ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{4} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= \frac{1}{4} ln tan (arcsec (\frac{4}{2} x)) + sec (arcsec (\frac{4}{2} x))

Simplify

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

= \frac{1}{4} ln 4 x^{2} - 1 + 2 x

Add a constant to the solution

= \frac{1}{4} ln 4 x^{2} - 1 + 2 x + C

\frac{\partial}{\partial x} (x^{2} + y^{2} + \frac{32}{x y})

\frac{d}{d x} (sec^{2} (2 x) 2)

x \to 1 lim (e^{(x \cdot 3 - x)})

x \to 1 - lim (x - 1)

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