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

Question

Evelyn Agard · Accepted Answer

#int1/((3+x)(1-x))dx=1/4ln(3+x)-1/4ln(1-x)# Let us first find partial fractions of #1/((3+x)(1-x))# and for this let #1/((3+x)(1-x))hArrA/(3+x)+B/(1-x)# or #1/((3+x)(1-x))hArr(A(1-x)+B(3+x))/((3+x)(1-x))# or #1/((3+x)(1-x))hArr((B-A)x+(A+3B))/((3+x)(1-x))# or Hence #B-A=0# or #A=B# and #A+3B=1#, i.e.#A=B=1/4# Hence #int1/((3+x)(1-x))dx=int[1/(4(3+x))+1/(4(1-x))]dx# or = #1/4int1/(3+x)dx+1/4int1/(1-x)dx# = #1/4ln(3+x)-1/4ln(1-x)#

Christopher Allers · Answer

undefined To integrate $ \frac{1}{{(3 + x)(1 - x)}} $ using partial fractions, follow these steps:

1. Express the given fraction as a sum of simpler fractions: 
$ \frac{1}{{(3 + x)(1 - x)}} = \frac{A}{3 + x} + \frac{B}{1 - x} $

2. Find the values of $ A $ and $ B $ by equating numerators:
$ 1 = A(1 - x) + B(3 + x) $

3. Solve for $ A $ and $ B $ by choosing suitable values of $ x $. Typically, choose $ x = 1 $ and $ x = -3 $ to simplify the equation.

4. Once you find the values of $ A $ and $ B $, rewrite the original fraction with the partial fraction decomposition.

5. Integrate each term separately using basic integration techniques.

6. Finally, combine the integrals to obtain the overall integral of the original expression.

7. The integral of $ \frac{1}{{(3 + x)(1 - x)}} $ using partial fractions will yield the solution.