Integral of -1/(2x)dx

Question

Evaluating the integral of -1/(2x)dx comes out as -1/2 ln|x| + c.

Using U-substitution for the same equation, the integral of -1/(2x)dx, comes out as -1/2 ln|-2x| + c.

Can anyone explain why these two methods result in different answers?

Thanks for the help.

Answer 1

-∫[1/(2x)]dx Let u = 2x. Then du = 2dx. So, dx = (1/2)du

So, we have -∫[1/u](1/2)du = -(1/2)ln l 2x l + C = -(1/2)[ln2 + lnlxl]+ C = -(1/2)ln lxl -(1/2)ln2 + C

Group all constants together and relabel as C to get -(1/2)lnl x l + C

Answer 2

∫(-1/(2x))dx

= (-1/2)∫(dx/x)

= (-1/2)ln(|x|)+C

Basically, we can factor out the constant (-1/2) from the integrand so we have a simple integral!

Answer 3

correct answer is -(1/2)ln|x| + c

integral of 1/2x dx = -1/2 intergral of 1/x= (-1/2)ln|x| + c

there is no need for u substitution.

you could let u=x but that's trivially the same problem

then get (-1/2)ln|u| + c, then replace u by x to get the same answer

-(1/2)ln|2x| = (-1/2)[ln2 + ln|x|] = -(1/2)ln2 -(1/2)ln|x|

which does not = -(1/2)ln|x| as -(1/2)ln2 does not =0

but both answers are true if you relabel c, as c+ any constant is still a constant

oddly similar to how you can add any finite constant to infinity and it's still infinity

another approach is check your answer

work the problem backwards

take the derivative of the integral and you should get back to -1/2x

then both -(1/2)ln|x| + c

and -(1/2)ln|2x|+ c

when differentiated will = -1/2x

they're both correct answers if you treat "c" as a different constant in the answers