How do you integrate $\int \frac{1}{r^{2} - 1} d r$ $int 1/(r^2-1)dr$ using partial fractions?

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How do you integrate $\int \frac{1}{r^{2} - 1} d r$ $int 1/(r^2-1)dr$ using partial fractions?

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Answer 1 · 2016-02-27T19:11:54

It is

$\int \frac{1}{r^{2} - 1} d r = \frac{1}{2} \cdot \int (\frac{1}{r - 1} - \frac{1}{r + 1}) d r = \frac{1}{2} \cdot (ln | r - 1 | - ln | r + 1 |) + c = \frac{1}{2} \cdot ln ∣ ∣ ∣ \frac{r - 1}{r + 1} ∣ ∣ ∣ + c$ $int 1/(r^2-1)dr=1/2*int (1/(r-1)-1/(r+1))dr=1/2*(lnabs(r-1)-lnabs(r+1))+c= 1/2*lnabs((r-1)/(r+1))+c$

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How do you integrate $\int \frac{1}{r^{2} - 1} d r$ $int 1/(r^2-1)dr$ using partial fractions?

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How do you integrate int 1/(r^2-1)dr∫1r2−1drint 1/(r^2-1)dr using partial fractions?

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How do you integrate $\int \frac{1}{r^{2} - 1} d r$ $int 1/(r^2-1)dr$ using partial fractions?