Question

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

Answer 1

`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)`