# How do you integrate # ln(x^2 -1)dx#?

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and by IBP

so now using partial fractions

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To integrate ln(x^2 -1)dx, you can use integration by parts. Let u = ln(x^2 - 1) and dv = dx. Then differentiate u to find du/dx, and integrate dv to find v.

u = ln(x^2 - 1) du/dx = 2x / (x^2 - 1) dv = dx v = x

Now apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, v, du, and dv into the formula:

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

Now integrate the second term using substitution method:

Let t = x^2 - 1 dt/dx = 2x dx = dt / (2x)

∫x * (2x / (x^2 - 1)) dx = ∫(2x^2 / (x^2 - 1)) dx = ∫(2t / t) * (dt / (2x)) = ∫(2t / t) * (dt / (2*√(t + 1))) = ∫dt / √(t + 1) = 2√(t + 1) + C = 2√(x^2) + C = 2|x| + C

Now substitute the integral back into the original equation:

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

where C is the constant of integration.

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