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integral of (x^2-25)/(x^2+4x-5)

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∫ x²−25x²+4x−5 dx

Solution

x−4ln|x−1|+C

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∫ x²−25x²+4x−5 dx

Take the partial fraction of x²−25x²+4x−5 : 1−4x−1

=∫ 1−4x−1 dx

Apply the Sum Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ 1dx−∫ 4x−1 dx

∫ 1dx=x

∫ 4x−1 dx=4ln|x−1|

=x−4ln|x−1|

Add a constant to the solution

=x−4ln|x−1|+C

Verify Result

=x−4ln|x−1|+C

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+▭▭	>	4	5	6	×	☐☐☐
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integral of (x^2-25)/(x^2+4x-5)

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integral of (x^2-25)/(x^2+4x-5)

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Solution steps

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