Question

Evaluate the following:

`int x^2/(1 - x^4) "d"x` put x² = t

Answer 1

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"`.