Question

Evaluate the following.

`int x/(4x^4 - 20x^2 - 3) dx`

Answer 1

I = `intx/(4x^4 - 20x^2 - 3)dx`

= `intx/(4[(x^2)^2 - 5x^2 - 3/4])dx`

= `1/4intx/((x^2)^2 - 5x^2 - 3/4)dx`

Put x² = t

Differentiate w.r.t. x both sides,

2x = `(dt)/(dx)`

2x.dx = dt

`x.dx = 1/2 dt`

I = `1/4int(1/2.dt)/(t^2 - 5t - 3/4)`

I = `1/4. 1/2int 1/(t^2 - 5t - 3/4)dt`

add and subtract `(1/2 xx -5)^2 = 25/4`

I = `1/8int1/((t^2 - 5t + 25/4) - 3/4 - 25/4)dt`

= `1/8int1/((t - 5/2)^2 - 7)dt`

= `1/8int1/((t - 5/2)^2 - (sqrt7)^2)dt`

I = `1/8. 1/(2(sqrt7)). log(|(t - 5/2 - sqrt7)/(t - 5/2 + sqrt7)|) + c`

= `1/(16sqrt7) . log|((2t)/2 - 5/2 - (2sqrt7)/2)/((2t)/2 - 5/2 + (2sqrt7)/2)| + c`

= `1/(16sqrt7) . log|(2t - 5 - 2sqrt7)/(2t - 5 + 2sqrt7)| + c`

I = `1/(16sqrt7) . log|(2x^2 - 5 - 2sqrt7)/(2x^2 - 5 + 2sqrt7)| + c`.

Answer 2

Let `I = int x/(4x^4 - 20x^2 - 3) dx`

Put x² = t

∴ 2x dx = dt

∴ `x  dx = dt/2`

∴ `I = int 1/(4t^2 - 20 t - 3) * dt/2`

= `1/2 xx 1/4 int 1/(t^2 - 5t - 3/4) dt`

= `1/8 int 1/((t^2 - 5t + 25/4) - 25/4 - 3/4) dt`

= `1/8 int 1/((t - 5/2)^2 - (sqrt 7)^2)`

= `1/8 xx 1/(2 sqrt 7) log |(t - 5/2 - sqrt 7)/(t - 5/2 + sqrt 7)| + c`

= `1/(16 sqrt 7) log |(2t - 5 - 2 sqrt 7)/(2t - 5 + 2 sqrt 7)| + c`

= `1/(16 sqrt 7) log |(2x^2 - 5 - 2 sqrt 7)/(2x^2 - 5 + 2 sqrt 7)| + c`.