Question

Integrate

Answer 1

\[\int\limits_{2}^{\infty} (n-4)/(n^2-2x+1)\]

Answer 2

with respect to x?

Answer 3

\[\int\limits\limits_{2}^{\infty} (\ln x)/\sqrt{x}\] dx

Answer 4

write n-4 as n-2 -2

write denominator as (n-2)^2
now split the terms as 1/n-2 - 2/ (n-2)^2

I hope you know how to integrate that..right?

Answer 5

Yes, steps please. I'm using these to find if the series converge with the integral test and I can't seem to integrate them.

Answer 6

Let me try the first one.

Answer 7

so is that supposed to be n^2-2n+1 then ??

Answer 8

I guess.

Answer 9

Sorry, typo.

Answer 10

Isn't
the denominator (x-1)^2?

If it is x-2, then I'd get an error.

Answer 11

\[2\int \frac{\ln \sqrt{x}}{\sqrt{x}}\]
Not sure but it might work, integrating with powers of e is easier than log.

Answer 12

yup it is. but when i split it becomes= (x-2)/ (x-2)^2 - 2/(x-2)^2

Answer 13

@Ishaan94 use by-parts for integrating that.

Answer 14

For the first one, substitute (Denominator)' = A(Numerator) + B, where A and B are constants.

Answer 15

But then it would become ln(x-2) + 3 / (x-2) for x to infinity. ln 0 doens't exist and 3/0 doesn't exist either.

Answer 16

I tried doing partial decomposition but it led to some decimal which is not right.

Answer 17

akshat, i think an easier way is using substitution
u = n-1
du = dn
\[\int\limits_{?}^{?}\frac{n-4}{(n-1)^{2}} = \frac{u-3}{u^{2}}\]

Answer 18

Yeah, I'd go with Dumbcow.

Answer 19

But how would I integrate u-3/u^2. Don't weneed a long integration by parts for that?

Answer 20

@dumbcow Substitution is advisable when a power is changed. Before substituting we still had a quadratic equation in denominator and we still have it after substituting.
You are not wrong but what you did right now is almost the same thing before substituting.

Answer 21

@hmm why don't you use the method I told you earlier?

write x-4 as x-2-2
split.
becomes (x-2)/(x-2)^2 - 2/(x-2)^2
= 1/(x-2) - 2/(x-2)^2

Now integrate! :)

Answer 22

Thank you for the first one. I got it, but I still can't figure out the second one. I tried to integrate it by parts but it gave me \[2\sqrt{x}\ln x -4\sqrt{x} from 1 \to infinity.\]
This gave me infniity minus infinity. I tried using the hopital rule but I end ud getting 0 times inifinity and such.

Answer 23

Second one doesn't converge

Answer 24

How do I prove it? I keep getting infinity and zero errors.

Answer 25

\[2\int \frac{\ln \sqrt{x}}{\sqrt{x}}\]
Put \(t = \ln \sqrt{x}\), I got \(4t e^t - 4e^t\), which clearly doesn't converge for \(x \to \infty \implies ln \sqrt{x} \to \infty \implies t \to \infty\)

Answer 26

\[\frac{u-3}{u^{2}} = \frac{1}{u} -\frac{3}{u^{2}}\]
after integrating
\[\ln u +\frac{3}{u}\]
\[=\ln(n-1) +\frac{3}{n-1}\]
evaluate from 2 to inf
\[= \infty\]

Answer 27

I don't understand how ln got transformed into e. Could I see the steps?

Answer 28

I got t/e^t when I plugged it in.

Answer 29

yes and then he used integration by parts

Answer 30

Now it becomes integration of 2 (e^t )* t dt
Now integrate it.