Question

Integrate the function:

`(sqrt(x^2 +1) [log(x^2 + 1) - 2log x])/x^4`

Answer 1

Let `I = int (sqrt(x^2 + 1)[log (x^2 + 1) - 2 log x])/x^4`dx

`= int (sqrt(x^2 + 1)[log (x^2 + 1) - log x^2])/x^4`dx

`= int sqrt(x^2 + 1)/x^4 * log ((x^2 + 1)/x^2)`dx

`= int (sqrt(x^2 + 1))/x^4 log (1 + 1/x^2)`dx

Putting x = tan θ,

⇒ dx = sec² θ dθ

∴ I = `sqrt(1 + tan^2 theta)/(tan^4 theta) log  (1 + 1/(tan^2  theta)) * sec^2 θ  dθ`

`= int (sec θ)/(tan^4 theta) * [log (1 + cot^2 theta)] sec^2 θ  d θ`

`= int [log (cosec^2 θ)] * (cos^4 θ)/(sin^4 θ) * sec^3 θ  dθ`

`= - 2 int (log sin θ) * (cos θ)/(sin^4 θ) dθ`

Put sin θ = t

cos θ dθ = dt

∴ I = `- 2 int (log t) * 1/t^4  dt`

Let us take log t as the first function.

I = `- 2 [(log t) int t^-4 dt - int (d/dt (log t) int t^-4 dt)dt]`

`= - 2 [log t(- 1/(3t^3)) - int 1/t(- 1/(3t^3))dt]`

`= -2 [- (log t)/3t^3 + 1/3 int t^-4 dt]`

`= 2/3 (log t)/t^3 - 2/3 (- 1/3 t^-3) + C`

`= 2/9 [(3 log t)/t^3 + 1/t^3] + C`

`= 2/9 [(3 log t + 1)/t^3] + C`

Now t = sin θ and tan θ = x

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

`therefore I = 2/9 [(3 log (x/(sqrt(x^2 + 1))) + 1)/(x/sqrt(1 + x^2))^3] + C`

`= (2 (1 + x))^(3/2)/(9x^3) [3 log  x/(sqrt(x^2 + 1)) + 1] + C`

`= 2/9 (1 + x^2)^(3/2)/x^3 * 3 log ((1 + x^2)/x^2)^(- 1/2) + 2/9 (1 + x^2)^(3/2)/x^3 + C`

`= - 1/3 (1 + 1/x^2)^(3/2) log (1 + 1/x^2) + 2/9 (1 + 1/x^2)^(3/2) + C`

`= - 1/3 (1 + 1/x^2)^(3/2) [log (1 + 1/x^2) - 2/3] + C`