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

1 Answer
May 13, 2016

#int1/((3+x)(1-x))dx=1/4ln(3+x)-1/4ln(1-x)#

Explanation:

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)#