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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2} d u

= \int \frac{sec ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (2 x)

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

Simplify

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

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

Add a constant to the solution

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

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\frac{d}{d x} (\frac{1 - sin ( x )}{1 + cos ( x )})

\frac{d}{d x} (x cos (y))

\int_{0}^{10} 2200 e^{0.024 x} - 1460 e^{0.018 x} d x

\int \frac{1}{T} d T

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