What is the integral of 1/(sqrt(9x^2-4)) ?

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

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integral of 1/(sqrt(9x^2-4))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{2 x + x ^{3}} d x

\int \frac{1}{x 64 x ^{4} - 1} d x

\frac{d}{d x} (\frac{1}{( x ^{3} - 2 x ^{2} + 7 ) ^{4}})

n = 1 \sum \infty 4 n - 1

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