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

Question

Samantha Adkison · Accepted Answer

#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# #x^3/(x^2-1)# #= (x^3-x+x)/(x^2-1)# #= (x(x^2-1)+x)/(x^2-1)# #= x + x/(x^2-1)# #= x + x/((x-1)(x+1))# #= x + A/(x-1) + B/(x+1)# #= x + (A(x+1)+B(x-1))/(x^2-1)# #= x + ((A+B)x + (A-B))/(x^2-1)# Equating coefficients we find: #{ (A+B = 1), (A-B = 0) :}# Hence: #A = B = 1/2# So: #x^3/(x^2-1) = x + 1/(2(x-1)) + 1/(2(x+1))# So: #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#

Scarlett Allison · Answer

undefined 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 $