Question

Integrate the function:

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

Answer 1

Let I = `int 1/(x^2(x^4 + 1)^(3/4)` dx

put x = `sqrt(tan theta)`

`therefore dx = 1/2 (tan theta)^(1/2 - 1) * sec^2 theta  d theta`

`= (sec^2 theta)/(2sqrt(tan theta))dθ`

`= int 1/(tan θ (tan^2 θ + 1)^(3/4)) * (sec^2 θ)/(2sqrt(tan θ))`dx

`= 1/2 int (sec^2 θ  d θ)/(tan θ * sec^(3/2) θ * sqrt(tan θ))`

`= 1/2 int (sec^2 θ dθ)/((tan θ * sec θ)^(3/2))`

`= 1/2 int (sec^2 θ  dθ)/((sin θ)/(cos θ) * sec θ)^(3/2)`

`= 1/2 int (sec^2 θ  dθ)/(sin^(3/2) θ (sec^2θ)^(3/2))`

`therefore 1/2 int (sec^2 θ  dθ)/(sin^(3/2) θ (sec^3 θ))`

`= 1/2 int (cos θ  dθ)/(sin^(3/2) θ)`

Putting sin θ = t,

cos θ dθ = dt

`therefore I = 1/2 int dt/t^(3//2) = 1/2 int t^(-3/2)` dt

`= 1/2 (-2)t^(- 1/2) + C`

`= - 1/sqrtt + C`

`= - 1/sqrt(sin theta) + C`

Putting x = `sqrt(tan θ)`,

tan θ = x²

`therefore sin theta = x^2/sqrt(1 + x^4)`

`sqrt(sin theta) = x/(1 + x^4)^(1/4)`

Putting this value in equation (1),

`I = - 1/sqrt(sin theta) + C`

`= ((1 + x^4)^(1/4))/x + C`

Hence, `int 1/(x^2(x^4 + 1)^(3/4)) dx `

`= - (1 + x^4)^(1//4)/x` + C

Answer 2

Let `I = int dx/(x^2 (x^4 + 1)^(3/4))`

`= dx/(x^5 (1 + 1/x^4)^(3/4))`

Put `1 + 1/x^4 = t`

⇒ `-4/x^5` dx = dt

∴ `I = -1/4 int dt/t^(3/4)`

`= -1/4 int^(-3/4)  dt`

`= -1/4 t^(1/4)/(1/4) + C`

`= -(t)^(1/4) + C`

`= - (1 + 1/x^4)^(1/4) + C`