integral of (x^4-2x^2+4x+1)/(x^3-x^2-x+1)

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∫ x⁴−2x²+4x+1x³−x²−x+1 dx

Solution

x²2 +x+ln|x−1|−2x−1 −ln|x+1|+C

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∫ x⁴−2x²+4x+1x³−x²−x+1 dx

Take the partial fraction of x⁴−2x²+4x+1x³−x²−x+1 : x+1+1x−1 +2(x−1)² −1x+1

=∫ x+1+1x−1 +2(x−1)² −1x+1 dx

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

=∫ xdx+∫ 1dx+∫ 1x−1 dx+∫ 2(x−1)² dx−∫ 1x+1 dx

∫ xdx=x²2

∫ 1dx=x

∫ 1x−1 dx=ln|x−1|

∫ 2(x−1)² dx=−2x−1

∫ 1x+1 dx=ln|x+1|

=x²2 +x+ln|x−1|−2x−1 −ln|x+1|

Add a constant to the solution

=x²2 +x+ln|x−1|−2x−1 −ln|x+1|+C

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
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integral of (x^4-2x^2+4x+1)/(x^3-x^2-x+1)

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