# How do you find #int (2x^2 -x)/((x^2 -1)^2)dx# using partial fractions?

First let's factor the denominator:

Now we see two repeated linear terms, so our partial fraction decomposition will look like this:

After getting common denominators you should get the following (Sorry, I kinda skipped a step):

This is a MESS to sort through, but a quick thing we can do is look at the coefficients in front of the powers on the left side...

now let's do the constants...

now we can finally rewrite the integral like so:

all together we have:

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To integrate ( \frac{2x^2 - x}{(x^2 - 1)^2} ) using partial fractions, we first factor the denominator ( (x^2 - 1)^2 ) as ( (x - 1)^2(x + 1)^2 ). Then, we express the integrand as a sum of partial fractions:

[ \frac{2x^2 - x}{(x - 1)^2(x + 1)^2} = \frac{A}{(x - 1)} + \frac{B}{(x - 1)^2} + \frac{C}{(x + 1)} + \frac{D}{(x + 1)^2} ]

Next, we find the values of ( A ), ( B ), ( C ), and ( D ) by multiplying both sides by the denominator ( (x - 1)^2(x + 1)^2 ) and then equating the numerators:

[ 2x^2 - x = A(x - 1)(x + 1)^2 + B(x + 1)^2 + C(x - 1)^2(x + 1) + D(x - 1)^2 ]

We can solve for ( A ), ( B ), ( C ), and ( D ) by substituting specific values of ( x ) that eliminate all but one of the terms on the right-hand side of the equation. After finding the values, we integrate each term separately.

[ A = 1 ] [ B = 1 ] [ C = -1 ] [ D = 1 ]

Now, the integral becomes:

[ \int \frac{1}{x - 1} + \frac{1}{(x - 1)^2} - \frac{1}{x + 1} + \frac{1}{(x + 1)^2} , dx ]

[ = \ln|x - 1| - \frac{1}{x - 1} - \ln|x + 1| - \frac{1}{x + 1} + C ]

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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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