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I have to calculate this integral: $$\int_{13}^\infty \frac{x+1}{x^3+x}$$ Now the indefinite integral of the integral above: $$\arctan(x)+\ln(x)-\frac{1}{2}\ln(x^2+1)+C$$ which I have to calculate, so I could calculate the definite integral, is correct. I checked it on Wolfram Alpha. However Wolfram is also telling me that the value of the my definite integral is around: 0.07972. And that's the problem. More below.

Here is the undefinite integral, modified so I can just put the values 13 and $\infty$ in and get the result: $$\lim_{x\to \infty}\left(\arctan(x)+\ln(x)-\frac{1}{2}\ln(x^2+1)\right)-\left(\arctan(13)+\ln(13)-\frac{1}{2}\ln(13^2+1)\right)$$ However In the left part of the integral above (the one with limit), when I put infinity into both logarithms there, I'll get $$\ln(\infty) - \frac{1}{2}\ln(\infty^2+1)=\infty - \infty.$$ But from what I know from school, that is undefined expression. I also noticed, that if I'll ignore the infinities created by logarithms in the integral below, I'll get the exact same number, that is acording to Wolfram, the value of my integral.

So is this some special case, where I can says that $\infty - \infty = 0$ or is there more to it? Thanks for answers.

And sorry if you had hard time reading this, English isn't my mother tongue.

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    $\begingroup$ MathJax hint: if you put a backslash before common functions you get the proper font and spacing, so \log (x) gives $\log (x)$ instead of log (x) which gives $log (x)$ $\endgroup$ – Ross Millikan Jan 6 '18 at 21:18
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    $\begingroup$ Don't "put infinity" in those logs. Better, use the properties of log to evaluate the limit... $\endgroup$ – DonAntonio Jan 6 '18 at 21:20
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The antiderivative is correct, and

$$\arctan x+\log\frac x{\sqrt{x^2+1}}=\left(\arctan x+\log\sqrt{\frac{x^2}{x^2+1}}\right)\xrightarrow[x\to\infty]{}\frac\pi2+\log1=\frac\pi2$$

Now add to that all the other constants $\;\arctan 13\;$ and etc.

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    $\begingroup$ I see. Thanks a lot. $\endgroup$ – Milmitt Jan 6 '18 at 23:45
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Intuitively, when $x$ gets very large, the $x^2+1$ is no different from $x^2$. You need to combine the two logs analytically to avoid the indeterminate form.

$\lim_{x \to \infty} \arctan (x)=\frac \pi 2$ is no problem. $$\lim_{x \to \infty} (\log(x)-\frac 12 \log(x^2+1))=\lim_{x \to \infty}(\log(x)-\log(\sqrt{x^2+1}))\\=\lim_{x \to \infty}\log \frac x{\sqrt{x^2+1}}\\=\lim_{x \to \infty}\log\frac 1{\sqrt{1+\frac 1{x^2}}}\\=0$$

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DonAntonio has given a correct answer, but I'd like to address what it seems to me your misunderstanding is.

Yes, $\infty-\infty$ is still undefined. The problem is that you can't always substitute into expressions to compute limits. In this case, as you've noticed, substituting $\infty$ into the expression produces an undefined expression. The correct approach in this situation is to use properties of logarithms to get $$\ln x - \frac{1}{2}\ln(x^2+1) = \ln\left(\frac{x}{\sqrt{x^2+1}}\right).$$ While we still can't substitute $\infty$ into the right hand expression, (because $\frac{\infty}{\infty}$ is undefined, we can divide numerator and denominator by $x$ to get $$\ln\left(\frac{x}{\sqrt{x^2+1}}\right)=\ln\left(\frac{1}{\sqrt{1+1/x^2}}\right).$$ Now we can substitute $\infty$ in to get $$\lim_{x\to\infty} \ln\left(\frac{x}{\sqrt{x^2+1}}\right)=\ln\left(\frac{1}{\sqrt{1+0}}\right)=\ln 1=0.$$

Now you can proceed as in DonAntonio's answer to get the final result.

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  • $\begingroup$ Thanks, I completely forgot that I can put logarithms together like that. $\endgroup$ – Milmitt Jan 6 '18 at 23:46
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You can rewrite the expression using logarithms properties as $\arctan x + \ln{\frac{x}{(x^2+1)^{1/2}}}$. We have: $$\ln{\frac{x}{(x^2+1)^{1/2}}}\sim_{+\infty}\ln{\frac{x}{(x^2)^{1/2}}}=\ln{1}=0.$$

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