Question

integral

Answer 1

i tried break it up, but the first integral, already doesnt converge

Answer 2

anyone able to help?

Answer 3

so don't break up

Answer 4

then im stuck lol

Answer 5

i cant integrate, the whole integral

Answer 6

1/ ln (b^2+1) - C ln (3x+1)
why not ?

Answer 7

sorry, lim b-> 0 1/ ln (b^2+1) - C [ln (3b+1)]

Answer 8

now do ln A - ln B = ... ?

Answer 9

lim b-> 0 1/ ln (b^2+1) - C/3 [ln (3b+1)]
= lim b-> 0 1/2 ln{ (b^2+1) / {C/3 [ln (3b+1)}]
= lim b-> 0 1/2 ln{ (b^2+1) / {[ln (3b+1)}^(C/3)]

Answer 10

getting that ?

Answer 11

b-> infinity...grrr making so many typos

Answer 12

what do i do now?

Answer 13

woops wrong one

Answer 14

what will be integral of c/3x+1 ?

Answer 15

what do i do now? how can i find \(C\)?

Answer 16

ln A - ln B is just ln (A/B)

Answer 17

not ln A/ln B

Answer 18

then does it equal
\[\lim_{b \rightarrow \infty} \ln ( \frac{(b^2+1)^2}{(3x+1)^{c/3}}\]??

Answer 19

yes

Answer 20

i still, dont know how to find \(C\)?

Answer 21

@amistre64 can you continue here ? i got to go, sorry.

Answer 22

ive got a test i have to study for, so im not able to focus that well on this stuff yet :)

Answer 23

okay

Answer 24

@.Sam. @thomaster @DebbieG ?

Answer 25

Here's what WolframAlpha got me for the integration:

Take the integral:
integral (x/(x^2+1)-C/(3 x+1)) dx
Integrate the sum term by term and factor out constants:
= integral x/(x^2+1) dx-C integral 1/(3 x+1) dx
For the integrand 1/(1+3 x), substitute u = 1+3 x and du = 3 dx:
= integral x/(x^2+1) dx-C/3 integral 1/u du
The integral of 1/u is log(u):
= integral x/(x^2+1) dx-1/3 C log(u)
For the integrand x/(1+x^2), substitute s = 1+x^2 and ds = 2 x dx:
= 1/2 integral 1/s ds-1/3 C log(u)
The integral of 1/s is log(s):
= (log(s))/2-1/3 C log(u)+constant
Substitute back for s = 1+x^2:
= 1/2 log(x^2+1)-1/3 C log(u)+constant
Substitute back for u = 1+3 x:
Answer: |
| = 1/2 log(x^2+1)-1/3 C log(3 x+1)+constant

But other than that you'll have to put in the limits yourself I think. And solve for C. Hope that helps. Sorry I can't do more. I'm doing homework myself.