The intgral of one over x^2-1

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To find

\displaystyle \int{{\frac{{dx}}{{{{x}^{2}}-1}}}}  

we do a perfect square factorisation to get

\displaystyle \int{{\frac{{dx}}{{{{x}^{2}}-1}}}}=\int{{\frac{{dx}}{{(x+1)(x-1)}}}}  

We can solve using partial fractions. We should find values A and B such that

\displaystyle \frac{1}{{(x+1)(x-1)}}=\frac{A}{{x+1}}+\frac{B}{{x-1}}  

We multiply all of this by the denominator of the LHS to get

\displaystyle 1=A(x-1)+B(x+1)  

Now, if we let x=1, we get 1=2B, or B=1/2. Now, let  x= –1 to get 1= –2A or A= –1/2. This gives us

\displaystyle \begin{array}{l}\int{{\frac{{dx}}{{{{x}^{2}}-1}}}}=\frac{1}{2}\int{{\left( {\frac{{-1}}{{x+1}}+\frac{1}{{x-1}}} \right)}}dx\\\quad \quad \quad =\frac{1}{2}\left( {-\ln \left| {x+1} \right|+\ln \left| {x-1} \right|} \right)+C\\\quad \quad \quad =\frac{1}{2}\ln \left| {\frac{{x-1}}{{x+1}}} \right|+C\end{array}  

The graph of the integrand and the answer look  like this:

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Next page : The integral of one over x^2+1Last modified: May 21, 2025 @ 12:13