Question

is the ln(-x) = -1/x dx?

Answer 1

\[\int\limits (-1/x) dx = - \ln|x|\]

Answer 2

\[\int\limits_{-3}^{-1}\frac{ du }{ x \sqrt{\ln \left| -x \right|} }\]

for this I get 2sqrt ln3 but the answer is -2sqrt ln 3

Answer 3

you'll need to break up the ln|-x|

Answer 4

for my u substitution I had u = ln(-x); du = -1/x; which gives me integral of -du/u

Answer 5

Should I break up the ln(-x) further?

Answer 6

so, you have ln|-x|, ln(-x)

ln|-x| = ln(-x) for -x > 0
ln|-x| = ln(x) for x <= 0

which case are you in?

Answer 7

I am going from -3 to -1 which means that I would be in the second case (ln|-x| = ln(x) for x <= 0) which means that ln(-x) = ln(x). Is this rigtht?

Answer 8

it means ln|-x| becomes ln(x).

Answer 9

I am slightly confused

Answer 10

so you have:

\[\int\limits \frac{dx}{x \ln(x)}\]

Answer 11

oops

Answer 12

\[\int\limits\limits \frac{dx}{x \sqrt{\ln(x)}}\]

Answer 13

u = ln(x)
du = (1/x) dx, so you have:

\[\int\limits \frac{du}{\sqrt{u}}\]

Answer 14

Did you take out the negative from both the integral and the ln to do that?

Answer 15

oh crap, I was dead wrong O.O Let's back up

ln|-x| = ln(-x) for -x > 0
ln|-x| = ln(x) for -x <= 0

so we're in the first case because -x > 0, means x < 0

so,
\[\int\limits \frac{dx}{x \sqrt{\ln(-x)}}\]

Answer 16

which means that I have u= ln|-x| du = 1/x and integral of du/u. Is this correct?

Answer 17

Basically what you are doing above is just splitting up the absolute value?

Answer 18

yes.

But you're supposed to let u = ln(-x), du = (1/x)dx

the integral becomes

Answer 19

Thanks

Answer 20

I got it now

Answer 21

which becomes 2 sqrt(ln(-x))

Answer 22

so you have 2 sqrt(ln(- -1)) - 2 sqrt(ln(- - 3)) = 2 [sqrt(ln(1)) - sqrt(ln(3))
= -2sqrt(ln(3))