You can reach integral of (x^2+1)/(x^2-1) answer on this page.
What is integral of $int frac{x^2+1}{x^2-1}dx$?
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| integral of (x^2+1)/(x^2-1) |
Solution:
Performing polynomial long division, we have that:
$int frac{x^2+1}{x^2-1}dx=int (1+frac{2}{x^2-1})dx$
$=int dx + int frac{2}{x^2-1}dx$
$=x+int frac{2}{x^2-1}dx$
Using partial fraction on the remaining integral, we get:
$frac{2}{x^2-1}=frac{A}{x-1}+frac{B}{x+1}=frac{A(x+1)+B(x-1)}{(x+1)(x-1)}=frac{(A+B)x+(A-B)}{x^2-1}$
Thus, A + B = 0 and A − B = 2. Adding the two equations together yields 2.A = 2, that is, A = 1, and B = − 1. So, we have that:
$int frac{2}{x^2-1}dx=int frac{1}{x-1}dx-int frac{1}{x+1}dx$
Therefore,
$int frac{x^2+1}{x^2-1}dx=x+int frac{2}{x^2-1}dx$
$=x+int frac{1}{x-1}dx-int frac{1}{x+1}dx$
$=x+ln|x-1|-ln|x+1|+C$
Answer :
$int frac{x^2+1}{x^2-1}dx=x+ln|x-1|-ln|x+1|+C$
