Metamath Proof Explorer


Theorem advlog

Description: The antiderivative of the logarithm. (Contributed by Mario Carneiro, 21-May-2016)

Ref Expression
Assertion advlog dx+xlogx1dx=x+logx

Proof

Step Hyp Ref Expression
1 reelprrecn
2 1 a1i
3 rpre x+x
4 3 adantl x+x
5 4 recnd x+x
6 1cnd x+1
7 recn xx
8 7 adantl xx
9 1red x1
10 2 dvmptid dxxdx=x1
11 rpssre +
12 11 a1i +
13 eqid TopOpenfld=TopOpenfld
14 13 tgioo2 topGenran.=TopOpenfld𝑡
15 ioorp 0+∞=+
16 iooretop 0+∞topGenran.
17 15 16 eqeltrri +topGenran.
18 17 a1i +topGenran.
19 2 8 9 10 12 14 13 18 dvmptres dx+xdx=x+1
20 relogcl x+logx
21 20 adantl x+logx
22 peano2rem logxlogx1
23 21 22 syl x+logx1
24 23 recnd x+logx1
25 rpreccl x+1x+
26 25 adantl x+1x+
27 26 rpcnd x+1x
28 21 recnd x+logx
29 relogf1o log+:+1-1 onto
30 f1of log+:+1-1 ontolog+:+
31 29 30 mp1i log+:+
32 31 feqmptd log+=x+log+x
33 fvres x+log+x=logx
34 33 mpteq2ia x+log+x=x+logx
35 32 34 eqtrdi log+=x+logx
36 35 oveq2d Dlog+=dx+logxdx
37 dvrelog Dlog+=x+1x
38 36 37 eqtr3di dx+logxdx=x+1x
39 0cnd x+0
40 1cnd x1
41 0cnd x0
42 1cnd 1
43 2 42 dvmptc dx1dx=x0
44 2 40 41 43 12 14 13 18 dvmptres dx+1dx=x+0
45 2 28 27 38 6 39 44 dvmptsub dx+logx1dx=x+1x0
46 27 subid1d x+1x0=1x
47 46 mpteq2dva x+1x0=x+1x
48 45 47 eqtrd dx+logx1dx=x+1x
49 2 5 6 19 24 27 48 dvmptmul dx+xlogx1dx=x+1logx1+1xx
50 24 mullidd x+1logx1=logx1
51 rpne0 x+x0
52 51 adantl x+x0
53 5 52 recid2d x+1xx=1
54 50 53 oveq12d x+1logx1+1xx=logx-1+1
55 ax-1cn 1
56 npcan logx1logx-1+1=logx
57 28 55 56 sylancl x+logx-1+1=logx
58 54 57 eqtrd x+1logx1+1xx=logx
59 58 mpteq2dva x+1logx1+1xx=x+logx
60 49 59 eqtrd dx+xlogx1dx=x+logx
61 60 mptru dx+xlogx1dx=x+logx