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Consider the following integrals.
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The first integral, 2 over y squared minus 1, can be evaluated using partial fraction decomposition.
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In this case, we would rewrite 2 over y squared minus 1 into the sum of partial fractions with denominators y minus 1 and y plus 1.
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And because they're both linear, their corresponding numerators are constants, let's call them a and.
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With this form, we can already evaluate the integral.
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For part b, we can use substitution here.
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We want to set the denominator y squared minus 1 equal to u, so that du over 2 is ydy.
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Then from there, the integral becomes the integral of one half du over u.
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You can also use partial fractions here.
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And if it's partial fraction, then y over y squared minus 1 is going to equal a over y minus 1 plus b over y plus 1...