Question

Improper Integration

can someone please enlighten me on how he got ln(2) from this integral ......? https://www.dropbox.com/s/frc1wjyt1hcwsmc/Screenshot%202016-04-03%2019.33.08.png?dl=0

Thanks a lot for reading

Answer 1

if I have 1/1-x^2 can I change the sign in the deno and pass a minus outside the integral so that I have 1/x^2 + 1 and then work with arctan ??

Answer 2

I haven't worked it out, but have you tried factoring it into partial fractions?

\[x^2-1 = (x-1)(x+1)\]

Answer 3

I tried using the arctangent

Answer 4

I got more or less the same result , but im not sure if I can do what I did to be able to work with arctangent

Answer 5

when I say the same what I really mean is something like 0.66 .... and ln(2) is something like 0.69 ..

Answer 6

I can see how to do it with partial fractions
you get
\[ \int_3^\infty \frac{1}{x-1}- \frac{1}{x+1} \ dx\]

Answer 7

@Christos Oh, you can't turn this into arctangent, since you will have to have a negative sign on one of the terms:

\[\frac{1}{x^2-1} = \frac{-1}{1-x^2}\]

Answer 8

You can use the tanh^-1(x)

Answer 9

I see :/

Answer 10

partial fractions -continued-
\[\int_3^\infty \ln(x-1) - \ln(x+1) = \ln \left(\frac{x-1}{x+2} \right) \bigg|_3^\infty\]

this is an improper integral, so you do the "limit" as x-> infinity for the upper bound
i.e. ln(1) =0

Answer 11

*(x-1)/(x+1) (not x+2 in the bottom)

Answer 12

the lower bound gives
- ln(2/4) = - ln(½) = ln(2)

Answer 13

I guess this is also partial fractions right ? https://www.dropbox.com/s/uaap7mf1jkln8y3/Screenshot%202016-04-03%2019.47.37.png?dl=0

Answer 14

I assume that is (ln x)^3 in the bottom?
off hand, I don't know...

Answer 15

ok

Answer 16

yeah its ln(x)^3

Answer 17

it looks like substitution
u= ln(x)^3

Answer 18

du = ln(x)3

Answer 19

seems like it was u substitution

Answer 20

d u^-2 = -2 u^-3 du = -2 (ln x)^-3 1/x dx

yes, substitution. Though I am stumbling over the details

Answer 21

although you have to jump to find 1/u^3

Answer 22

thx a lot