Question

Does the integral of 1/(x(1+x)) from 1 to infinity converge or diverge?

Answer 1

converge since the denominator continuously increases and hence the function is continuously decreasing.

Answer 2

do you know what it converges to? i got something like -.6931 but i dont think thats right

Answer 3

\[\int_1^\infty\frac{1}{x(1+x)}dx\]

Answer 4

let \[\frac{A}{x}+\frac{B}{x+1}=\frac{1}{x(x+1)}\]

Answer 5

simplifying, we can get
\[A(x+1)+B(x)=1\\
A=1\qquad B=-1\]

Answer 6

\[
I=\int_1^\infty\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\\
I=[\ln x-\ln (x+1)]_1^\infty\\
I=\left[\ln\frac{x}{x+1}\right]_1^\infty\\
I=\left[\ln\frac{1}{1+{1\over x}}\right]_1^\infty\\
I=\ln\frac{1}{1+0}-\ln\frac{1}{1+1}\\
I=\ln(1)-\ln(2^{-1})\\
I=\ln(2)
\]

Answer 7

oh! okay and ln(2)=.6931 which is what i got! thank you very much!

Answer 8

notice that it is not negative...

Answer 9

so, how did you get the decimal number? did you do the same integration?
also, if you did, how did you preceed to do the integration without knowing if it converges?

Answer 10

well i did the same thing you did but left it as ln(x)-ln(1+x) and with some silly algebra i got the negative. i got infinity minus infinity at some point which confused me and i second guessed if it converged

Answer 11

ok.

Answer 12

thank you for your help