any one know the integration of logx

Answer 1

int logx = xlogx -x +c

check the answer

(xlogx-x+c)'= x(1/x)+logx(1)-1+0= logx

Answer 2

For the area beneath y=ln(x) to x-axis

see Mark's answer, and use definite integral.

from x=1 to x=e the area =[e*1-e]-[1*0-1]=1 sq unit.

from x=1 to arbitrary x>1 the area =[xlnx-x]-[1*0-1]=xlx-x+1+1,,,kind of cool, like 'rectangle-x" plus 1

For area beneath y=log₁₀ (x) to x-axis

see Doug's answer and use definite integral

from x=1 to x=10 the area =[10*1-(10/ln10)]-[1*0-(1/ln10)]=10-(9/ln(10))

Answer 3

If you mean lnx, use integration by parts with u = lnx and dv = dx.

Then, du = (1/x)dx and v = ∫dv = x.

So, ∫ lnxdx = ∫udv = uv - ∫vdu = xlnx - ∫dx = xlnx - x + C

Answer 4

Assuming the post really intends common log.

Integration by parts.

u = log x

dv = ∫dx

du = 1/(xln(10)) dx

v = x

I = xlogx - ∫x(1/xln10)dx

I = xlogx - x/ln(10) + C

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