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## What is integral of $\int \frac{x^2+1}{x^2-1}dx$?

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 :**

**$\int \frac{x^2+1}{x^2-1}dx=x+ln|x-1|-ln|x+1|+C$**

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