What is the integral of 1/(sqrt(81x^2-1296)) ?

The integral of 1/(sqrt(81x^2-1296)) is 1/9 ln((|sqrt(x^2-16)+x|)/4)+C

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integral of 1/(sqrt(81x^2-1296))

\int \frac{1}{81 x ^{2} - 1296} d x

\frac{1}{9} ln (\frac{x ^{2} - 16 + x}{4}) + C

Solution steps

\int \frac{1}{81 x ^{2} - 1296} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{9} d u

= \int \frac{sec ( u )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{9} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{9}{36} x)

= \frac{1}{9} ln tan (arcsec (\frac{9}{36} x)) + sec (arcsec (\frac{9}{36} x))

Simplify

\frac{1}{9} ln tan (arcsec (\frac{9}{36} x)) + sec (arcsec (\frac{9}{36} x)) : \frac{1}{9} ln (\frac{x ^{2} - 16 + x}{4})

= \frac{1}{9} ln (\frac{x ^{2} - 16 + x}{4})

Add a constant to the solution

= \frac{1}{9} ln (\frac{x ^{2} - 16 + x}{4}) + C

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What is the integral of 1/(sqrt(81x^2-1296)) ?
The integral of 1/(sqrt(81x^2-1296)) is 1/9 ln((|sqrt(x^2-16)+x|)/4)+C