The Student Room Group
Reply 1
Both are but the C values will be different.
cvat
how come when i integrate it by doing 0.5 x integral of 1/x

i get a different answer to just integrating it without taking the factor of a half,

so for the first one i get 1/2 ln(x)

and for the second one i get 1/2 ln (2x)

(i know when i differentiate both of them, i get the same result, but which is the correct integral?)


They are both correct. The only difference is that in each case, the constant of integration will be different. Obviously this doesn't matter if you're substituting in limits, as the constant cancels out anyway - you'll get the same answer for both.
Reply 3
its 1/2 ln x
Reply 4
cvat
how come when i integrate it by doing 0.5 x integral of 1/x

i get a different answer to just integrating it without taking the factor of a half,

so for the first one i get 1/2 ln(x)

and for the second one i get 1/2 ln (2x)

(i know when i differentiate both of them, i get the same result, but which is the correct integral?)


As the above poster said, both are correct, but as a general rule, if have an integral 1kx  dx\displaystyle\int \frac{1}{kx} \; dx I find that you should just leave it as 1klnx+C\frac{1}{k} \ln x + C. But that's just me...
Reply 5
so if it wasnt a definite integral, and it asked me that question, i'd get the marks for either of them?
Reply 6
cvat
so if it wasnt a definite integral, and it asked me that question, i'd get the marks for either of them?


Yes.
Reply 7
ok thanks :smile:, but yh i think 1/2 ln x + C, does look better too

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