What is the integral of ln(x^2-4) ?

The integral of ln(x^2-4) is xln(x^2-4)+4(1/2 ln| x/2+1|-1/2 ln| x/2-1|-x/2)+C

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integral of ln(x^2-4)

\int ln (x^{2} - 4) d x

x ln (x^{2} - 4) + 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + C

Solution steps

\int ln (x^{2} - 4) d x

Apply Integration By Parts

: x ln (x^{2} - 4) - \int \frac{2 x ^{2}}{x ^{2} - 4} d x

= x ln (x^{2} - 4) - \int \frac{2 x ^{2}}{x ^{2} - 4} d x

\int \frac{2 x ^{2}}{x ^{2} - 4} d x = - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2})

= x ln (x^{2} - 4) - (- 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}))

Simplify

= x ln (x^{2} - 4) + 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2})

Add a constant to the solution

= x ln (x^{2} - 4) + 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + C

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What is the integral of ln(x^2-4) ?
The integral of ln(x^2-4) is xln(x^2-4)+4(1/2 ln| x/2+1|-1/2 ln| x/2-1|-x/2)+C