What is the integral of 2/(sqrt(4x^2-9)) ?

The integral of 2/(sqrt(4x^2-9)) is ln(1/3 |sqrt(4x^2-9)+2x|)+C

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integral of 2/(sqrt(4x^2-9))

\int \frac{2}{4 x ^{2} - 9} d x

ln (\frac{1}{3} 4 x^{2} - 9 + 2 x) + C

Solution steps

\int \frac{2}{4 x ^{2} - 9} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 x ^{2} - 9} d x

Apply Trigonometric Substitution

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

= 2 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = arcsec (\frac{2}{3} x)

= 2 \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x))

Simplify

2 \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{3} x)) + sec (arcsec (\frac{2}{3} x)) : ln (\frac{1}{3} 4 x^{2} - 9 + 2 x)

= ln (\frac{1}{3} 4 x^{2} - 9 + 2 x)

Add a constant to the solution

= ln (\frac{1}{3} 4 x^{2} - 9 + 2 x) + C

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What is the integral of 2/(sqrt(4x^2-9)) ?
The integral of 2/(sqrt(4x^2-9)) is ln(1/3 |sqrt(4x^2-9)+2x|)+C