I built a column of integrals from the CRC Handbook for Chemistry and Physics 69 edition that celebrated 75 years starting on page A-20.
In[]:=
Inactive[Integrate][
n
x
,x]
Out[]=
∫
n
x
x
In[]:=
Inactive[Integrate][
n
x
,x]==Activate[Inactive[Integrate][
n
x
,x]]
Out[]=
∫
n
x
x
1+n
x
1+n
In[]:=
FormatIndefiniteIntegral//ClearAll​​FormatIndefiniteIntegral[input_,variable_]:=Inactive[Integrate][input,variable]==FullSimplify@Activate[Inactive[Integrate][input,variable]]
In[]:=
FormatIndefiniteIntegral[
n
x
,n]
Out[]=
∫
n
x
n
n
x
Log[x]
In[]:=
FormatIndefiniteIntegral[Exp[x],x]
Out[]=
∫
x

x
x

In[]:=
FormatIndefiniteIntegral
4
x
2
a
-
2
x
,x
Out[]=

4
x
2
a
-
2
x
x-
1
8
x
2
a
-
2
x
(3
2
a
+2
2
x
)+
3
8

4
a
Log-x+
2
a
-
2
x

I will create an integral table based on the CRC Handbook 69 edition.
In[]:=
inputs=a,af[x],(p+q),
n
x
,
f'[x]
f[x]
,
1
x
,
f'[x]
2
f[x]
,Exp[x],Exp[ax],Power[b,ax],Log[x],
x
a
Log[x],
1
2
a
+
2
x
,
1
2
a
-
2
x
,
1
2
x
-
2
a
,
1
2
a
-
2
x
,
1
2
x
+
2
a
,
1
2
x
-
2
a
,
1
x
2
x
-
2
a
,
1
x
2
a
+
2
x
,
1
x
2
a
-
2
x
,
n
(a+bx)
,x
n
(a+bx)
,
2
x
n
(a+bx)
,
1
a+bx
,
1
2
(a+bx)
,
1
3
(a+bx)
,
x
a+bx
,
x
2
(a+bx)
,
x
n
(a+bx)
,
2
x
a+bx
,
2
x
2
(a+bx)
,
2
x
3
(a+bx)
,
2
x
n
(a+bx)
,
1
x(a+bx)
,
1
x
2
(a+bx)
,
1
x
3
(a+bx)
,
1
2
x
(a+bx)
,
1
3
x
(a+bx)
,
1
2
x
2
(a+bx)
(*endofsectiononformscontaininga+bx*),(*startofsectiononformscontaining
2
c
±
2
x
,
2
x
-
2
c
*)
1
2
c
+
2
x
,
1
2
c
-
2
x
,
1
2
x
-
2
c
,
x
2
c
+
2
x
,
x
2
c
-
2
x
(*thebooklists
x
2
c
±
2
x
,butIhavetoentertwoseparateintegralsfor
x
2
c
+
2
x
and
x
2
c
-
2
x
*),
x
n+1
(
2
c
+
2
x
)
,
x
n+1
(
2
c
-
2
x
)
,
1
n
(
2
c
+
2
x
)
,
x
n+1
(
2
c
-
2
x
)
,
1
n
(
2
x
-
2
c
)
,
x
1
(
2
x
-
2
c
)
,
x
n+1
(
2
x
-
2
c
)
;
In[]:=
Table[FormatIndefiniteIntegral[input,x],{input,inputs}]
In[]:=
Column[Table[FormatIndefiniteIntegral[input,x],{input,inputs}]]
Out[]=
In[]:=
Column[Table[FormatIndefiniteIntegral[input,x],{input,inputs}]]//TraditionalForm
Out[]//TraditionalForm=
I’m going to apply my function to questions from the MIT Integration Bee.
In[]:=
MITIntegrationBeeQualtifyingExamIntegrals=<|"2022-1"->
1+Cos[x]
x+Sin[x]
,"2022-4"->
Sinh[x]
Cosh[x]-Sinh[x]
,"2022-5"->
x
x-1
+
x+1
,"2022-7"->
3
x
Sin[
2
x
],"2022-8"->
x
1-
4
x
,"2022-9"->
1
2
Cosh[x]
,"2022-13"->
3
x
1+x+
2
x
2
+
3
x
6
,"2022-14"->Sin[x+Sin[x]]-Sin[x-Sin[x]],"2022-15"->
4
Tan[x]
3
Sec[x]
+
2
Tan[x]
5
Sec[x]
,"2022-16"->(1+Log[x])Log[Log[x]],"2022-17"->
1
1+Sin[x]
+
1
1+Cos[x]
+
1
1+Tan[x]
+
1
1+Cot[x]
+
1
1+Sec[x]
+
1
1+Csc[x]
,"2022-18"->
1
x-
2
x
,"2022-20"->
1
1+
2
Cos[x]
,"2020-1"->
Log[2x]
xLog[x]
,"2020-5"->
1
2
x
+
2
(x-1)
|>;
In[]:=
Column[Table[FormatIndefiniteIntegral[input,x],{input,Values[MITIntegrationBeeQualtifyingExamIntegrals]}]]//TraditionalForm
Out[]//TraditionalForm=

cos(x)+1
x+sin(x)
xlog(x+sin(x))

sinh(x)
cosh(x)-sinh(x)
x
1
4
(-2x+
2x

+1)

x
x-1
+
x+1
x
1
15

x-1
(-3
2
x
+x+2)+(3x-2)
3/2
(x+1)

∫
3
x
sin(
2
x
)x
1
2
(sin(
2
x
)-
2
x
cos(
2
x
))

x
1-
4
x
x
1
2
-1
tanh
(
2
x
)
∫
2
sech
(x)xtanh(x)

3
x
3
x
6
+
2
x
2
+x+1
x6(x-log(x(x(x+3)+6)+6))
∫(sin(x+sin(x))-sin(x-sin(x)))x-2cos(sin(x))
∫(
2
tan
(x)
5
sec
(x)+
4
tan
(x)
3
sec
(x))x
1
3
3
tan
(x)
3
sec
(x)
∫(log(x)+1)log(log(x))xx(log(x)log(log(x))-1)

1
sin(x)+1
+
1
cos(x)+1
+
1
tan(x)+1
+
1
cot(x)+1
+
1
csc(x)+1
+
1
sec(x)+1
x3x

1
x-
2
x
x
2
3/2
(x-1)
log(
x-1
-
x
)
3/2
(1-x)

1
2
cos
(x)+1
x
-1
cot
(
2
cot(x))
2

log(2x)
xlog(x)
xlog(x)+log(2)log(log(x))

1
2
x
+
2
(x-1)
x
-1
tan
(2(x-1)+1)