How do you integrate $f (x) = \frac{x^{2} + 1}{(x^{2} - 1) (x - 4)}$ $f(x)=(x^2+1)/((x^2-1)(x-4))$ using partial fractions?

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How do you integrate $f (x) = \frac{x^{2} + 1}{(x^{2} - 1) (x - 4)}$ $f(x)=(x^2+1)/((x^2-1)(x-4))$ using partial fractions?

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Answer 1 · 2016-12-05T18:22:05

$- \frac{1}{3} ln (x - 1) + \frac{1}{5} ln (x + 1) + \frac{17}{15} ln (x - 4) + C$ $-1/3 ln (x-1) +1/5 ln (x+1) +17/15 ln (x-4) +C$ (constant of integration)

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How do you integrate $f (x) = \frac{x^{2} + 1}{(x^{2} - 1) (x - 4)}$ $f(x)=(x^2+1)/((x^2-1)(x-4))$ using partial fractions?

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How do you integrate f(x)=(x^2+1)/((x^2-1)(x-4))f(x)=x2+1(x2−1)(x−4)f(x)=(x^2+1)/((x^2-1)(x-4)) using partial fractions?

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How do you integrate $f (x) = \frac{x^{2} + 1}{(x^{2} - 1) (x - 4)}$ $f(x)=(x^2+1)/((x^2-1)(x-4))$ using partial fractions?