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Question
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
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Solution
Put x2 = t
Then 2x dx = dt.
Now I = `int (x^3"d"x)/(x^4 + 3x^2 + 2)`
= `1/2 int "tdt"/("t"^2 + 3"t" + 2)`
Consider `"t"/("t"^2 + 3"t" + 2) = "A"/("t" + 1) + "B"/("t" + 2)`
Comparing coefficient, we get A = –1, B = 2.
Then I = `1/2[2 int "dt"/("t" + 2) - int "dt"/("t" + 1)]`
= `1/2 [2log|"t" + 2| - log|"t" + 1|]`
= `log|(x^2 + 2)/sqrt(x^2 + 1)| + "C"`
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