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Question
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
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Solution
We have I = `int (x^2 + x)/(x^4 - 9) "d"x`
= `int x^2/(x^4 - 9) "d"x + (x"d"x)/(x^4 - 9)`
= I1 + I2
Now I1 = int x^3/(x^4 - 9)`
Put t = x4 – 9
So that 4x3 dx = dt.
Therefore I1 = `1/4 int "dt"/"t"`
= `1/4 log|"t"| + "C"_1`
= `1/4 log|x^4 - 9| + "C"_1`
Again, I2 = `int (x"d"x)/(x^4 - 9)`
Put x2 = u
So that 2x dx = du
Then I2 = `1/2 int "du"/("u"^2 - (3)^2)`
= `1/(2 xx 6) log|("u" - 3)/("u" + 3)| + "C"_2`
= `1/12 log|(x^2 - 3)/(x^2 + 3)| + "C"_2`.
Thus I = I1 + I2
= `1/4 log|x^4 - 9| + 1/12 log|(x^2 - 3)/(x^2 + 3)| + "C"`
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