How do you integrate $int 1/ [(x-1) ( x+ 1) ^2]$ using partial fractions?

Question

1317 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-13T07:08:47

$int1/((x-1)(x+1)^2)dx$

= $1/4ln(x-1)-1/4ln(x+1)+1/(2(x+1))+c$

Let us first find partial fractions of $1/((x-1)(x+1)^2)$ and for this let

$1/((x-1)(x+1)^2)hArrA/(x-1)+B/(x+1)+C/(x+1)^2$ or

$1/((x-1)(x+1)^2)hArr(A(x+1)^2+B(x-1)(x+1)+C(x-1))/((x-1)(x+1)^2)$ or

$1/((x-1)(x+1)^2)hArr(A(x^2+2x+1)+B(x^2-1)+C(x-1))/((x-1)(x+1)^2)$ or

$1/((x-1)(x+1)^2)hArr((A+B)x^2+(2A+C)x+(A-B-C))/((x-1)(x+1)^2)$

Therefore $A+B=0$ and $2A+C=0$ and $A-B-C=1$

From these $B=-A$ , $C=-2A$ , Hence $A-(-A)-(-2A)=1$ or $4A=1$

Hence $A=1/4$ , $B=-1/4$ and $C=-1/2$

Hence $int1/((x-1)(x+1)^2)dx$

= $int[1/(4(x-1))-1/(4(x+1))-1/(2(x+1)^2)]dx$

= $1/4ln(x-1)-1/4ln(x+1)+1/(2(x+1))+c$

How do you integrate int 1/ [(x-1) ( x+ 1) ^2] int 1/ [(x-1) ( x+ 1) ^2] using partial fractions?