I need the method to evaluate this integral (the closed-form if possible).

∫π/201+tanhx1+tanxdx

I used the relationship between tanx and tanhx but it didn’t work. Any help?

**Answer**

I=∫π/201+tanh(x)1+tan(x)dx=π4+∫π/20tanh(x)1+tan(x)dx

I1=∫π/20tanh(x)1+tan(x)dx

tanh(x)=sinh(x)cosh(x)=ex−e−xex+e−x=1−2e−xex+e−x

so:

I1=π4−2∫π/2011+tan(x)e−xex+e−xdx

I2=∫π/2011+tan(x)e−xex+e−xdx

e−xex+e−x=e−2x1−(−e−2x)=e−2x∞∑n=0(−1)ne−2nx=∞∑n=0(−1)ne−2(n+1)x

and so:

I2=∫π/2011+tan(x)∞∑n=0(−1)ne−2(n+1)xdx=∞∑n=0(−1)n∫π/20e−2(n+1)x1+tan(x)dx

as for this integral its quite messy and I’m not sure what to do from here, It would be easier for 0 to π/4 I think. I will say that as n increases the terms get smaller very quickly so an approximation of the first few would be quite accurate if possible.

One possible way I have noticed is that:

e−2.5(n+1)x≤e−2(n+1)x1+tan(x)≤e−2.4(n+1)x

so if:

J(a)=∫π/20e−axdx=1−e−aπ/2a

so we have:

∞∑n=0(−1)n1−e−2.5(n+1)π/22.5(n+1)≤I2≤∞∑n=0(−1)n1−e−2.4(n+1)π/22.4(n+1)

and according to wolfram alpha these sums converge and arent too ugly, and we know that:

I=π/2−2I2

thats the best I can do at the moment I’ll take another look sometime. It is also worth noting that:

∞∑n=0(−1)nn+1=ln(2)

and the second part of the sum could be expanded into a double summation

Back to add a little to this answer, so far we know:

ln22.5+12.5∞∑n=11ne−2.5nπ/2≤I2≤ln22.4+12.4∞∑n=11ne−2.4nπ/2

we will try and focus on sums of the form:

S(α)=∞∑n=1exp(−αn)n=−ln(e−α(eα−1))

according to wolfram alpha, which we are able to simplify to:

S(α)=α−ln(eα−1)

S(α)α=1−ln(eα−1)α

**Attribution***Source : Link , Question Author : E.H.E , Answer Author : Henry Lee*