Question

Integrate the following with respect to x.

`x/(2x^4 - 3x^2 - 2)`

Answer 1

`x/(2x^4 - 3x^2 - 2) = x/((2x^2 + 1)(x^2 - 2))`

Let x² = t

Then 2x dx = dt

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

= `1/2 int "dt"/((2"t" + 1)("t" - 2))`

We make use of partial fraction method

Let `1/((2"t" + 1)("t" - 2)) = "A"/(2"t" + 1) + "B"/("t" - 2)`

1 = A(t − 2) + B(2t + 1)

Let t = 2, Then 1 = 5B

⇒ B = `1/5`

Let t = `- 1/2`,Then 1 = `(-5)/2`A

⇒ A = `(- 2)/5`

So `1/2 int "dt"/((2"t" + 1)("t" - 2)) = 1/2 int ((-2)/5)/(2"t"+ 1)  "dt" + 1/2 int (1/5)/("t" - 2) "dt"`

= `(- log|2"t" + 1|)/(5(2)) + 1/10 log |"t" - 2| + "c"`

Putting t = x², we get = `(- log|2x^2 + 1|)/10 + 1/10 log|x^2 - 2| + "c"`

= `1/10 log |(x^2 - 2)/(2x^2 + 1)|  + "c"`

Using log (a) − log (b) = `log "a"/"b"`