How do you integrate #[(x^2 + 1)/(x^2 - 1)]dx#?

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Answer 1 · 2016-06-20T06:28:09

#int[(x^2+1)/(x^2-1)]dx=x-ln(x+1)+ln(x-1)+c#

#int[(x^2+1)/(x^2-1)]dx#

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

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

= #x+int2/(x^2-1)dx#

As #(x^2-1)=(x+1)(x-1)#, let #2/((x+1)(x-1))hArrA/(x+1)+B/(x-1)# i.e.

#2/((x+1)(x-1))hArr(A(x-1)+B(x+1))/((x+1)(x-1))# or

#2/((x+1)(x-1))hArr((A+B)x-A+B)/((x+1)(x-1))#

i.e. #A+B=0# and #-A+B=2# and adding them we get #2B=2#

or #B=1# and #A=-1#. Hence

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

= #x-ln(x+1)+ln(x-1)+c#