# How to solve this improper integral? ?

Hi,

How to solve

### 1 Answer

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- Ian HLv 71 month ago
Let u = tanh(𝛑x) = (e^2𝛑x – 1)/(e^2𝛑x + 1) = 1 - 2/(e^2𝛑x + 1)

du/dx = 4𝛑e^2𝛑x/(e^2𝛑x + 1)^2

∫udx = ∫tanh(𝛑x)dx = x + ln/(e^2𝛑x + 1) – 2𝛑x = ln/(e^2𝛑x + 1) – 𝛑x

Let v = 1/(16x^3 + x) = 1/x – 16(16x^2 + 1)

dv/dx = -(48x^2 + 1)’[x^2(16x^2 + 1)^2]

∫v dx = ln(x) – (1/2)ln(16x^2 + 1)

Use integration by parts formula for I = ∫u(x)*v(x) dx

https://www.mathsisfun.com/calculus/integration-by...

I = u∫v dx −∫[du/dx * (∫v dx)] dx

Applying limits zero to infinity should simplify that result

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