# How do you integrate #x^3/ (x^2 -1)# using partial fractions?

#int x^3/(x^2-1) dx = int (x + 1/(2(x-1)) + 1/(2(x+1))) dx#

#=1/2(x^2 + ln(abs(x-1)) + ln(abs(x+1))) + C#

Equating coefficients we find:

Hence:

So:

So:

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To integrate ( \frac{x^3}{x^2 - 1} ) using partial fractions, first, we factor the denominator as ( (x - 1)(x + 1) ). Then, we express ( \frac{x^3}{x^2 - 1} ) as the sum of two fractions with unknown constants:

( \frac{x^3}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} )

Next, we clear the denominators by multiplying both sides of the equation by ( (x - 1)(x + 1) ):

( x^3 = A(x + 1) + B(x - 1) )

Expanding and combining like terms:

( x^3 = Ax + A + Bx - B )

Now, we equate coefficients:

For ( x^3 ) term: ( A + B = 0 )

For ( x^1 ) term: ( A + B = 0 )

For ( x^0 ) term: ( A - B = 1 )

Solving this system of equations gives ( A = \frac{1}{2} ) and ( B = -\frac{1}{2} ).

Now, we can rewrite the original integral as:

( \int \frac{x^3}{x^2 - 1} , dx = \int \left( \frac{\frac{1}{2}}{x - 1} - \frac{\frac{1}{2}}{x + 1} \right) , dx )

Then, we integrate each term separately:

( \int \frac{\frac{1}{2}}{x - 1} , dx = \frac{1}{2} \ln|x - 1| + C_1 )

( \int \frac{\frac{-1}{2}}{x + 1} , dx = -\frac{1}{2} \ln|x + 1| + C_2 )

So, the final result is:

( \int \frac{x^3}{x^2 - 1} , dx = \frac{1}{2} \ln|x - 1| - \frac{1}{2} \ln|x + 1| + C )

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