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

The integral of-1/(9xsqrt(81x^2-9)) is -1/27 arcsec(3x)+C

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integral of-1/(9xsqrt(81x^2-9))

\int - \frac{1}{9 x 81 x ^{2} - 9} d x

- \frac{1}{27} arcsec (3 x) + C

Solution steps

\int - \frac{1}{9 x 81 x ^{2} - 9} d x

Take the constant out:

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

= - \frac{1}{9} \cdot \int \frac{1}{x 81 x ^{2} - 9} d x

Apply Trigonometric Substitution

= - \frac{1}{9} \cdot \int \frac{1}{3} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - \frac{1}{9} \cdot \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{9}{3} x)

= - \frac{1}{9} \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{9}{3} x)

Simplify

- \frac{1}{9} \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{9}{3} x) : - \frac{1}{27} arcsec (3 x)

= - \frac{1}{27} arcsec (3 x)

Add a constant to the solution

= - \frac{1}{27} arcsec (3 x) + C

What is the integral of-1/(9xsqrt(81x^2-9)) ?
The integral of-1/(9xsqrt(81x^2-9)) is -1/27 arcsec(3x)+C