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

Answer 1

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

#xln(x^2-1)-2x-ln|x-1|+ln|x+1|+C#

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

We start by using integration by parts.

Let #u = ln(x^2-1)# and #dv = dx#
Then #du = (2x)/(x^2-1)dx# and #v = x#
By the integration by parts formula #intudv = uv - intvdu#
#intln(x^2-1)dx = xln(x^2-1) - int(2x^2)/(x^2-1)dx#
#=xln(x^2-1)-2int(1+1/(x^2-1))dx#
#=xln(x^2-1)-2int1dx -2int1/(x^2-1)dx#
#=xln(x^2-1)-2x-2int1/((x+1)(x-1))dx#

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

#1/((x+1)(x-1))=A/(x+1)+B/(x-1)#
#=> 1 = A(x-1) + B(x+1)#
#=(A+B)x + (-A+B)#
#=>{(A+B=0), (-A+B=1):}#
#=>{(A = -1/2),(B=1/2):}#
#=>int(1/((x+1)(x-1))dx = int((1/2)/(x-1)-(1/2)/(x+1))dx#
#=1/2(int1/(x-1)dx-int1/(x+1)dx)#
#=1/2(ln|x-1|-ln|x+1|)+C#

Substituting this back into the prior result, we have

#intln(x^2-1)dx = # #= xln(x^2-1)-2x-ln|x-1|+ln|x+1|+C#
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Answer 2

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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