integration

let y = logx with base a,
To find out ∫y dx, let's see what dy/dx is first....

for y = logx with base a,
then a^y =x
yIna = Inx
y = Inx / Ina
dy/dx = 1/(xIna)

Now, to find out ∫y dx, use Intergation by Part...
assume the base is 10,

∫logx dx
= xlogx - ∫x d(logx)
= xlogx - ∫x [1/(xIn10)] dx
= xlogx - [1/In10]x + C


Checking :
let y = xlogx - [1/In10]x + C
dy/dx = logx + x[ 1/(xIn10)] - 1/In10
dy/dx = logx

∫logxdx
=xlogx-∫x(1/x)dx,by integration by parts
=xlogx-∫dx
=xlogx-x+C