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

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

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

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

ln (\frac{1}{9} x^{2} - 81 + x) + C

Solution steps

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

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

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

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{1}{9} x)

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

Simplify

ln tan (arcsec (\frac{1}{9} x)) + sec (arcsec (\frac{1}{9} x)) : ln (\frac{1}{9} x^{2} - 81 + x)

= ln (\frac{1}{9} x^{2} - 81 + x)

Add a constant to the solution

= ln (\frac{1}{9} x^{2} - 81 + x) + C

y^{^{'}} = - \frac{3 y}{t} - 2 - t^{- 4}, y (1) = 4

\frac{\partial}{\partial y} (\frac{x - 3}{y ^{2} + 1})

x \to - \infty lim (x - 5)

\int_{- 2}^{2} 32 - 8 y^{2} d y

\int_{6}^{8} \frac{46}{( x - 6 ) ^{3}} d x

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