Question

Evaluate the following:

`int (x^2"d"x)/(x^4 - x^2 - 12)`

Answer 1

Let I = `int (x^2"d"x)/(x^4 - x^2 - 12)`

= `int x^2/(x^4 - 4x^2 + 3x^2 - 12) "d"x`

= `int x^2/(x^2(x^2 - 4) + 3(x^2 - 4)) "d"x`

= `int x^2/((x^2 - 4)(x^2 + 3)) "d"x`

Put x² = t for the purpose of partial fraction.

We get `"t"/(("t" - 4)("t" + 3))`

Let `"t"/(("t" - 4)("t" + 3)) = "A"/("t" - 4) + "B"/("t" + 3)` .....[where A and B are arbitrary constants]

`"t"/(("t" - 4)("t" + 3)) = ("A"("t" + 3) + "B"("t" - 4))/(("t" - 4)("t" + 3))`

⇒ t = At + 3A + Bt – 4B

Comparing the like terms, we get

A + B = 1 and 3A – 4B = 0

⇒ 3A = 4B

∴ A = `4/3 "B"`

Now `4/3 "B" + "B"` = 1

`7/3 "B"` = 1

∴ B = `3/7` and A = `4/3 xx 3/7 = 4/7`

So, A = `4/7` and B = `3/7`

∴ `int x^2/((x^2 - 4)(x^2 + 3)) "d"x`

= `4/7 int 1/(x^2 - 4)  "d"x + 3/7 int 1/(x^2 + 3)  "d"x`

= `4/7 int 1/(x^2 - (2)^2) "d"x + 3/7 int 1/(x^2 + (sqrt(3)^2)  "d"x`

= `4/7 xx 1/(2 xx 2) log|(x - 2)/(x + 2)| + 3/7 xx 1/sqrt(3) tan^-1  x/sqrt(3)`

= `1/7 log |(x - 2)/(x + 2)| + sqrt(3)/7 tan^-1 x/sqrt(3) + "C"`

Hence, I = `1/7 log |(x - 2)/(x + 2)| + sqrt(3)/7 tan^-1  x/sqrt(3) + "C"`.