Re: the example from W|A, I think Step 2’s verbiage could be improved by clarifying that the substitution, u=ex applies only to the radical sub-expression, not the entire integrand. I would be curious to learn how Mathematica’s code handles it.​​It applies to the entire integral.​​In step 2, perhaps this might help:
∫
t

2t

+4
t
can be rewritten as:​
∫
2
(
t

)
+4
(
t

t
)The first parenthesized expression becomes u and the second parenthesized expression becomes u, resulting in:​
∫
2
u
+4
u

In[]:=
∫Sqrt[E^(2t)+4]E^tt
Indefinite integrals
Approximate form
Hide steps
∫
2t

+4
t

t
1
2
t

2t

+4
+2
-1
sinh
t

2
+constant
Possible intermediate steps
Take the integral:
∫
t

2t

+4
t
For the integrand
t

2t

+4
, substitute u
t

and u
t

t:
 ∫
2
u
+4
u
For the integrand
2
u
+4
, substitute u2tan(s) and u2
2
sec
(s)s. Then
2
u
+4

4
2
tan
(s)+4
 2sec(s) and s
-1
tan
u
2
:
 2∫2
3
sec
(s)s
Factor out constants:
 4∫
3
sec
(s)s
Use the reduction formula, ∫
m
sec
(s)s =
sin(s)
m-1
sec
(s)
m-1
+
m-2
m-1
∫
-2+m
sec
(s)s, where m3:
 2tan(s)sec(s)+2∫sec(s)s
The integral of sec(s) is log(tan(s)+sec(s)):
 2tan(s)sec(s)+2log(tan(s)+sec(s))+constant
Substitute back for s
-1
tan
u
2
:
 2tan
-1
tan
u
2
sec
-1
tan
u
2
+2logtan
-1
tan
u
2
+sec
-1
tan
u
2
+constant
Simplify using sec(
-1
tan
(z))
2
z
+1
and tan(
-1
tan
(z))z:

1
2
2
u
+4
u+2log
1
2

2
u
+4
+u+constant
Substitute back for u
t

:

1
2
t

2t

+4
+2log
1
2

2t

+4
+
t

+constant
Which is equivalent for restricted t values to:
Answer:

1
2
t

2t

+4
+2
-1
sinh
t

2
+constant
Plots of the integral
min
max
min
max
Alternate forms of the integral
1
2
t

2t

+4
+4
-1
sinh
t

2
+constant
1
2
t

2t

+4
+2log
2t

4
+1
+
t

2
+constant
log(x) is the natural logarithm »
Series expansion of the integral at t = 0
1
2
5
+4
-1
sinh
1
2
+
5
t+
3
2
t
5
+
22
3
t
15
5
+
17
4
t
25
5
+O(
5
t
)
(Taylor series)
Big-O notation »
Definite integral (mean square over a period)
More digits
2π
∫
0
2t

(4+
2t

)t≈2.2239154962×
-15
10
…
Definite integral over a half-period
More digits
π
∫
0
t

4+
2t

t≈-4.1609152777…