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Question
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
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Solution
Let x2 = t.
Then `x^2/(x^4 + x^2 - 2) = "t"/("t"^2 + "t" - 2)`
= `"t"/(("t" + 2)("t" - 1))`
= `"A"/("t" + 2) + "B"/("t" - 1)`
So t = A(t – 1) + B(t + 2)
Comparing coefficients, we get A = `2/3`, B = `1/3`.
So `x^2/(x^4 + x^2 - 2) = 2/3 1/(x^2 + 2) + 1/3 1/(x^2 - 1)`
Therefore, `int x^2/(x^4 + x^2 - 2) "d"x`
= `2/3 int 1/(x^2 + 2) "d"x + 1/3 int "dx"/(x^2 - 1)`
= `2/3 1/sqrt(2) tan^-1 x/sqrt(2) + 1/6 log |(x + 1)/(x + 1)| + "C"`
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