With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that
![I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx](https://tex.z-dn.net/?f=I%20%3D%20%5Cdisplaystyle%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7B%5Csqrt%5B%5Cphi%5D%7Bx%7D%20%5Ctan%5E%7B-1%7D%28x%29%7D%7B%281%2Bx%5E%5Cphi%29%5E2%7D%20%5C%2C%20dx%20%3D%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bx%5E%7B%5Cphi-1%7D%20%5Ctan%5E%7B-1%7D%28x%29%7D%7Bx%20%281%2Bx%5E%5Cphi%29%5E2%7D%20%5C%2C%20dx)
Replace 
:
Split the integral at x = 1. For the integral over [1, ∞), substitute 
:
The integrals involving tan⁻¹ disappear, and we're left with
Substitution:
2x + (6(1/2x - 6)) = 19
2x + 3x - 36 = 19
5x - 36 = 19
+ 36
5x = 55
÷ 5
x = 11
y = (1/2 × 11) - 6
y = 5.5 - 6
y = -0.5
Elimination:
y = 1/2x - 6
- y
0 = 1/2x - 6 - y
+ 6
1/2x - y = 6
3x - 6y = 36
2x + 6y = 19
(add)
5x = 55
÷ 5
x = 11
y = (1/2 × 11) - 6
y = 5.5 - 6
y = -0.5
I hope this helps! Let me know if you need me to explain why I did some things :)
X increases by one
y increases by 2
(5, 1) - > (x + 1, y + 2) -> (6, 3)
Answer:
plz mark me as brsinliest..
