Question

Find the integral of (1/x-4x/(1+4x^2)) from 1 to infinity.

Answer 1

looks like 2 log functions to me

Answer 2

I did this problem and got infinity minus (0-(1/2)ln(5)) but the answer should be (1/2)ln(5/4), not (1/2)ln(5).

Answer 3

\[\lim_{a \rightarrow \infty} \int\limits_{1}^{a}(\frac{1}{x}-\frac{4x}{1+4x^2}) dx\]?

Answer 4

Right, but at the end, I got (1/2)ln(5) instead of (1/2)ln(5/4), isn't that weird?

Answer 5

did you get 2 log expressions like campbell said?

Answer 6

Yes, I got [ln(x)-(1/2)ln(1+4x^2)] and evaluated from 1 to infinity and I got (infinity-infinity)-(0-(1/2)ln(5)) and I got (1/2)ln(5). But that's not the right answer. I supposed to get (1/2)ln(5/4).

Answer 7

\[\lim_{a \rightarrow \infty}(\ln|x|-\frac{1}{2} \ln|1+4x^2|)|_{1}^{a}\]

\[\lim_{a \rightarrow \infty}[(\ln|a|-\frac{1}{2}|1+4a^2|)-(\ln|1|-\frac{1}{2}\ln|1+4(1)^2|)]\]


\[\lim_{a \rightarrow \infty}\ln|\frac{a}{\sqrt{1+4a^2}}|-0+\frac{1}{2}\ln|5| \]

you need to right that one part as one expression you can't say inf-inf=0

Answer 8

write*

Answer 9

Got it.