What is the integral of ln(sqrt(x^2-9)) ?

The integral of ln(sqrt(x^2-9)) is 1/2 (xln(x^2-9)+6(1/2 ln| x/3+1|-1/2 ln| x/3-1|-x/3))+C

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integral of ln(sqrt(x^2-9))

\int ln (x^{2} - 9) d x

\frac{1}{2} (x ln (x^{2} - 9) + 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})) + C

Solution steps

\int ln (x^{2} - 9) d x

Simplify

ln (x^{2} - 9) : \frac{1}{2} ln (x^{2} - 9)

= \int \frac{1}{2} ln (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}{2} \cdot \int ln (x^{2} - 9) d x

Apply Integration By Parts

: x ln (x^{2} - 9) - \int \frac{2 x ^{2}}{x ^{2} - 9} d x

= \frac{1}{2} (x ln (x^{2} - 9) - \int \frac{2 x ^{2}}{x ^{2} - 9} d x)

\int \frac{2 x ^{2}}{x ^{2} - 9} d x = - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})

= \frac{1}{2} (x ln (x^{2} - 9) - (- 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})))

Simplify

= \frac{1}{2} (x ln (x^{2} - 9) + 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}))

Add a constant to the solution

= \frac{1}{2} (x ln (x^{2} - 9) + 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})) + C

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What is the integral of ln(sqrt(x^2-9)) ?
The integral of ln(sqrt(x^2-9)) is 1/2 (xln(x^2-9)+6(1/2 ln| x/3+1|-1/2 ln| x/3-1|-x/3))+C