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

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Answer 1 · 2016-05-13T12:49:50

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

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

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

= $1/4ln(3+x)-1/4ln(1-x)$

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