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

The integral of 1/(xsqrt(x^2-81)) is 1/9 arcsec(1/9 x)+C

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

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

\frac{1}{9} arcsec (\frac{1}{9} x) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{9} d u

= \int \frac{1}{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 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{9} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

= \frac{1}{9} arcsec (\frac{1}{9} x) + C

f (t) = t^{4} - 20 t^{3} + 112 t^{2}

f (x) = arccos (x) + x, at x = 0

t \frac{d y}{d t} + 18 y = \frac{1}{t}

f (x) = \frac{2 x - 1}{x + 3}, at x = 3

\int - sin (t) d t

What is the integral of 1/(xsqrt(x^2-81)) ?
The integral of 1/(xsqrt(x^2-81)) is 1/9 arcsec(1/9 x)+C