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

The integral of ln(x^2-9) is 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(x^2-9)

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

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

Apply Integration By Parts

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

= 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})

= 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

= 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

= 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

\frac{\partial ^{3}}{\partial x \partial y \partial z} (e^{x y z^{8}})

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\frac{d}{d x} (cos (arcsin (x)))

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6 x^{2}, 0, 1, 5

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