I need help to this problem!!

Question

Integral of tan inverse of x divided by x, limits are from 1/2014 to 2014

Answer 1

If we recognize a symmetry of sorts around 1, we can try the substitution x = 1/u.  This gives us that dx = -1/u² du:

 

∫tan^-1(x)/x dx = ∫tan^-1(1/u)/(1/u) (-1/u²) du = -∫(π/2 - tan^-1(u))/u du

 

Now, if we consider the limits of integration, the last integral is from 2014 to 1/2014, which means we can swap the limits and change the sign.  Apologize that this format doesn't give me a clear way of expressing that, but with both definite integrals evaluated over 1/2014 to 2014, we get:

 

∫tan^-1(x)/x dx = ∫(π/2 - tan^-1(u))/u du = ∫(π/2u)du - ∫tan^-1(u)/u du

 

∫tan^-1(x)/x dx + ∫tan^-1(u)/u du = ∫(π/2u)du = π/2 [ln(2014) - ln(1/2014)] = π/2 [2 ln(2014)]

 

Since the definite integrals on the left are equal, we can add them and divide by 2 to get:

 

∫tan^-1(x)/x dx = π/2 ln(2014) ≈ 11.95

Answer 2

∫(tan^-1x)/x dx  

 

 

If we use integration by parts, we have

uv-∫vdu

u = tan^-1x                 dv = (1/x) dx

du = 1/(1+x²) dx      v = lnx

 

This gives us

 

ln(x) tan^-1x - ∫(ln x)/(1+x²) dx 

 

for the integral part, we need to use integration by parts again.

u = lnx                 dv = (1+x²) dx

du = (1/x) dx        v = x + x³/3 

 

ln(x) tan^-1x - [lnx (x+x³/3) - ∫(x+x³/3)/x dx]

ln(x) tan^-1x - lnx (x+x³/3) + ∫(1+x²/3) dx

ln(x) tan^-1x - lnx (x+x³/3) - (x+x³/9)

 

I'll let you apply the limits