# How do you evaluate the integral of #intln(x^2 - 1)dx#?

Use integration by parts and integration by partial fractions to find that the integral evaluates to

We will proceed using integration by parts and partial fraction decomposition .

We start by using integration by parts.

To evaluate the remaining integral, we can use partial fractions.

Substituting this back into the prior result, we have

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To evaluate the integral ∫ ln(x^2 - 1) dx, we can use integration by parts. Let u = ln(x^2 - 1) and dv = dx. Then, du = (1 / (x^2 - 1)) * 2x dx and v = x.

Using the integration by parts formula ∫ u dv = uv - ∫ v du, we get:

∫ ln(x^2 - 1) dx = x ln(x^2 - 1) - ∫ x * (1 / (x^2 - 1)) * 2x dx

This simplifies to:

∫ ln(x^2 - 1) dx = x ln(x^2 - 1) - 2∫(x^2 / (x^2 - 1)) dx

Now, we need to tackle the integral ∫(x^2 / (x^2 - 1)) dx. This can be done by partial fraction decomposition. Decomposing (x^2 / (x^2 - 1)) into partial fractions, we get:

x^2 / (x^2 - 1) = 1 + 1 / (x^2 - 1)

The integral becomes:

∫ (x^2 / (x^2 - 1)) dx = ∫ (1 + 1 / (x^2 - 1)) dx

Now, integrate each term separately:

∫ dx + ∫ (1 / (x^2 - 1)) dx

The integral of dx is simply x. For the second integral, we can use the substitution method. Letting u = x^2 - 1, we have du = 2x dx. Thus:

∫ (1 / (x^2 - 1)) dx = (1/2) ∫ (1 / u) du = (1/2) ln|u| + C

Substituting back u = x^2 - 1, we get:

(1/2) ln|x^2 - 1| + C

Now, combining all the parts together, we have:

∫ ln(x^2 - 1) dx = x ln(x^2 - 1) - 2x - (1/2) ln|x^2 - 1| + C

Where C is the constant of integration.

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