Log integration (1 Viewer)

[rama_v]

Can I please get some help on this integration question:

If y = (lnx)/x , find dy/dx and hence show that [Int from e^2 to e] (1-lnx)/xlnx dx = ln2 -1

I can find dy/dx but I dont know where to go from there. (btw this is a 3 unit question.)

 

[FinalFantasy]

Can I please get some help on this integration question:

If y = (lnx)/x , find dy/dx and hence show that [Int from e^2 to e] (1-lnx)/xlnx dx = ln2 -1

I can find dy/dx but I dont know where to go from there. (btw this is a 3 unit question.)

y = (lnx)/x
dy\dx=[(1\x)*x-lnx]\x²=(1-lnx)\x²

[Int from e^2 to e] (1-lnx)/xlnx dx
=int. 1\xlnx dx-int. 1\x dx both with same terminals
let I=int. 1\xlnx dx
let u=lnx, du=1\x dx
terminals are now from 2 to 1
.: I=int. u^-1 du=ln |u|=ln2-ln1=ln2

[Int from e^2 to e] (1-lnx)/xlnx dx
=ln2-ln x(with terminals e^2 to e)
=ln2-(2-1)=ln2-1

btw when u say "Int from e^2 to e", it means the e² is at the bottom.. so u should say from e to e²(unless im wrong)

 

[rama_v]

y = (lnx)/x
dy\dx=[(1\x)*x-lnx]\x²=(1-lnx)\x²

[Int from e^2 to e] (1-lnx)/xlnx dx
=int. 1\xlnx dx-int. 1\x dx both with same terminals
let I=int. 1\xlnx dx
let u=lnx, du=1\x dx
terminals are now from 2 to 1
.: I=int. u^-1 du=ln |u|=ln2-ln1=ln2

[Int from e^2 to e] (1-lnx)/xlnx dx
=ln2-ln x(with terminals e^2 to e)
=ln2-(2-1)=ln2-1

btw when u say "Int from e^2 to e", it means the e² is at the bottom.. so u should say from e to e²(unless im wrong)

Thanks ooh and yeah your right i should have said form e to e^2, my mistake.