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

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

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integral of 1/(sqrt((2x-1)(2x+1)))

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

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

Solution steps

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

Multiply by the conjugate of

2 x - 1 : \frac{4 x ^{2} - 1}{2 x + 1}

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

Multiply

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

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

Apply Trigonometric Substitution

= \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

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