Complicated integration problem

Evaluate the definite integral
.....2
.....∫ (e^x) / ( 1 + e ^(2x) ) dx
....0

THANKS

for a problem like this one, use an suitable substitution. Let's try u = e^x
so
du = e^x dx
when x = 0, u = 1
when x = 2, u = e²

.....2
.....∫ (e^x) / ( 1 + e ^(2x) ) dx
....0
=
...e²
....∫ du / ( 1 + u²)
....1
Now we need to do a trig substitution

u = tan(v)
du = sec²(v) dv
if u = 1 then v = tan⁻ֿ¹(1) = π/4
if u = e² then v = tan⁻ֿ¹(e²)

...tan⁻ֿ¹(e²)
....∫ sec²(v) dv / ( 1 + tan²(v))
....π/4

=
...tan⁻ֿ¹(e²)
....∫ sec²(v) dv / sec²(v)
....π/4
=
...tan⁻ֿ¹(e²)
....∫ dv
....π/4

= v [from π/4 to tan⁻ֿ¹(e²)]

= tan⁻ֿ¹(e²) - π/4

That was a good one!

If you are using integration tables then you can go from :

...e²
....∫ du / ( 1 + u²)
....1
to
= tan⁻ֿ¹(u) from 1 to e²

= tan⁻ֿ¹(e²) - π/4

I would start with the substitution
u = e^x
This means du = e^x dx and e^(2x) = u²

∫ (e^x) / (1 + e^(2x) dx becomes ∫ du / (1 + u²)

Now I would use the substitution
u = tan v
This means du = sec² v dv

The integration now becomes
∫ (sec² v) / (1 + tan² v) dv
which becomes
∫ (sec² v) / (sec² v) dv
which becomes
∫ dv
which gives v
which is tan^(-1) u
which is tan^(-1) e^x

Now put in the limits to evaluate

.....2
.....∫ (e˟) / ( 1 + e ²˟ ) dx
....0

Let u = e˟ ==> du = e˟ dx

When x=0 --> u = 1
When x= 2 --> u = e²

So you have

.....e²
.....∫ du/ ( 1 + u ² )
....1

= tanˉ¹ (e²) - tanˉ¹ (1) = tanˉ¹ (e²) - π/4 ≈ 0.65088

u=e^x
du=e^x dx

and you get

∫ du / ( 1 + u^2 )

where new limits of integration are from 1 to e^2/(1+e^4)

so you just get

arctan[e^2/(1+e^4)] - arctan[1]

not sure but may be dx = pi/4~~0.785398