Question

find the value of the constant C for which the integral
from 0 to infinity of (x/(x^2+1)-C/(3x+1)) dx converges. evaluate the integral for this value of C.

Answer 1

you might want to review the condition for an expansion to be converging. i.e. the convergence test

Answer 2

after integrate, i got this 1/6 ln (x^2+1)^3/(3x+1)^2C . then, how can i get the value of C ?

Answer 3

i think you should simplify the thing up there first and check for convergence.
though since ln1=0, maybe this would work: (x^2+1)^3=(3x+1)^2C

Answer 4

maybe you confusd with my answer .. after integrate , i got \[\frac{ 1 }{ 6 }\ln[ \frac{ (x^{2} + 1)^{3}}{ (3x + 1)^{2C} } ]\]

Answer 5

then.. how to get the value of C ?

Answer 6

You forgot this: "the integral from 0 to infinity"

So it's \(\Large\displaystyle \left. \dfrac{1}{6}\ln \left|\dfrac{(x^2 - 1)^3}{(3x+1)^{2C}}\right|\right|_{0}^{\infty}\)

\(= \large\displaystyle\lim_{x \rightarrow\infty}\dfrac{1}{6}\ln \left|\dfrac{(x^2 - 1)^3}{(3x+1)^{2C}}\right| - \dfrac{1}{6}\ln \left|\dfrac{((0)^2 - 1)^3}{(3(0)+1)^{2C}}\right|\)

Second part coverges so we only have to focus on the first one.

\( \large\displaystyle\lim_{x \rightarrow\infty}\dfrac{1}{6}\ln \left|\dfrac{(x^2 - 1)^3}{(3x+1)^{2C}}\right|= \large\displaystyle\lim_{x \rightarrow\infty}\dfrac{1}{6}\ln \left|\dfrac{x^6 \cdots}{(9x^2 \cdots)^{C}}\right|\)

Take a closer look, in order for this limit to converges, degree in denominator need to be equal or greater than degree in numerator so \(\Large\boxed{C\ge 3}\)

Answer 7

Is this clear? No?

Answer 8

Whoops fatal mistake. C can only be equal to 3. If C is greater than 3, then we eventually will get ln(0) which is undefined.

Answer 9

okay.. thankyou, i got that. :) then how to evaluate the integral with the value of C? can i just substitute the value C=3 in the original eq ?

Answer 10

Yeah.

Answer 11

but after subs with C=3, im having trouble with the final answer. i got the final ans \[\frac{ 1 }{ 2 } \ln \frac{ \infty }{ \infty } \]

Answer 12

Use L'Hopital's Rule.

Answer 13

i cant get the answer..

Answer 14

\(\large\displaystyle\lim_{x \rightarrow\infty}\dfrac{1}{6}\ln \left|\dfrac{(x^2 - 1)^3}{(3x+1)^{6}}\right| = \large\displaystyle\lim_{x \rightarrow\infty}\dfrac{1}{6}\ln \left|\dfrac{x^6 \cdots}{(3^6x^6\cdots}\right|\)
If you keeping use l'Hopital Rules, 6 times, you eventually should get
\[= \large\displaystyle\lim_{x \rightarrow\infty}\dfrac{1}{6}\ln \left|\dfrac{6!}{3^6 6!}\right| = \large\displaystyle\lim_{x \rightarrow\infty}\dfrac{1}{6}\ln \left|\dfrac{1}{3^6}\right| = \dfrac{1}{6}\ln \left(\dfrac{1}{3^6}\right) = \boxed{\ln \left(\dfrac{1}{3}\right)}\]

Answer 15

Or -ln(3)

Answer 16

Instead of literally take L'Hopital's rule several times, you can just cancel everything, leave only coefficient of term with highest power.

Answer 17

how did you get \[\ln \frac{ 1 }{ 3 }\]

Answer 18

okay yess. got that..sorry, so the answer is \[\ln \frac{ 1 }{ 3 }\] ?

Answer 19

This is what I get, so yes.

Answer 20

okay.. thanks for the help, ;)

Answer 21

where r you from ?

Answer 22

You're welcome. I'm from Texas, why?

Answer 23

ohh nothing. by the way, thanks ya (: