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Question
Evaluate the following:
`int x^2/(1 - x^4) "d"x` put x2 = t
Solution
Let I = `int x^2/(1 - x^4) "d"x`
= `int x^2/((1 - x^2)(1 + x^2)) "d"x`
Put x2 = t for the purpose of partial fractions.
We get `"t"/((1 - "t")(1 + "t"))`
Resolving into partial fractions we put
`"t"/((1 - "t")(1 + "t")) = "A"/(1 - "t") + "B"/(1 + "t")` .....[where A and B are arbitrary constants]
⇒ `"t"/((1 - "t")(1 + "t")) = ("A"(1 + "t") + "B"(1 - "t"))/((1 - "t")(1 + "t"))`
⇒ t = A + At + B – Bt
Comparing the like terms, we get A – B = 1 and A + B = 0
Solving the above equations
We have A = `1/2` and B = `- 1/2`
∴ I = `int (1/2)/(1 - x^2) "d"x + int ((-1)/2)/(1 + x^2) "d"x` ...(Putting t = x2)
= `1/2 * 1/(2*1) log |(1 + x)/(1 - x)| - 1/2 tan^-1x + "C"`
= `1/4 log |(1 + x)/(1 - x)| - 1/2 tan^-1x + 'C"`
Hence, I = `1/4 log |(1 + x)/(1 - x)| - 1/2 tan^-1x + "C"`.
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