Help me to show $\int_1^\infty \frac{1}{x+x^3}dx = \frac{\ln 2}{2}$? I have this exercise, and I get the right result. But while I think the first part is ok, the second part is from a formal stand point pretty hairy. So here the first part (I left some step out, but they should be for the most of you obvious):
$$\int_1^\infty \frac{1}{x+x^3} \,dx = \int_1^\infty \frac{1}{x(1+x^2)} \,dx= \int_1^\infty \frac{1}{x}-\frac{x}{(1+x^2)} \,dx= \left[\ln(x)\right]_1^\infty - \left[\frac{1}{2} ln(1+u)\right]_1^\infty$$
In the following second part, at least when I start to do arithmetic manipulation with $\infty$, it's not formal anymore.
$$ \left[\ln(x)\right]_1^\infty - \left[\frac{1}{2} ln(1-u)\right]_1^\infty 
= \ln(\infty)-ln(1)-\frac{1}{2} \ln(1-\infty)+\frac{1}{2}\ln(2)
= \infty -\frac{1}{2} \infty +\frac{1}{2}\ln(2)
= \frac{\ln 2}{2}$$
So it would be great if someone could show me a formal way to solve this integral.
 A: Substitute $x = \tan{\theta}$, $dx = \sec^2{\theta} d \theta$:
$$\int_1^{\infty} \frac{dx}{x (1+x^2)} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \: \cot{\theta} = \left [ \log{\sin{\theta}} \right ]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\log{\frac{1}{\sqrt{2}}} $$
A: An improper integral is simply a limit of proper integrals:
$$\int_1^\infty\frac1{x+x^3}dx=\lim_{a\to\infty}\int_1^a\left(\frac1x-\frac{x}{1+x^2}\right)dx\;.$$
Keep the limits until you’re done finding the antiderivatives, and keep the pieces together:
$$\begin{align*}
\int_1^\infty\frac1{x+x^3}dx&=\lim_{a\to\infty}\int_1^a\left(\frac1x-\frac{x}{1+x^2}\right)dx\\\\
&=\lim_{a\to\infty}\left[\ln x-\frac12\ln\left(1+x^2\right)\right]_1^a\\\\
&=\lim_{a\to\infty}\left[\ln x-\ln\left(1+x^2\right)^{1/2}\right]_1^a\\\\
&=\lim_{a\to\infty}\left[\ln\frac{x}{\left(1+x^2\right)^{1/2}}\right]_1^a\\\\
&=\lim_{a\to\infty}\ln\frac{a}{\left(1+a^2\right)^{1/2}}-\ln 2^{-1/2}\\\\
&=\frac12\ln 2\;,
\end{align*}$$
since $\dfrac{a}{\left(1+a^2\right)^{1/2}}\to 1$ as $a\to\infty$.
A: Set $x = \dfrac1u$. Hence, $dx = - \dfrac{du}{u^2}$. This gives us
$$I = \int_1^{\infty} \dfrac{dx}{x+x^3} = \int_0^1 \dfrac{du}{u^2 \left(\dfrac1u + \dfrac1{u^3} \right)} = \int_0^1 \dfrac{u du}{1+u^2} = \left. \dfrac12 \log(1+u^2) \right \vert_{u=0}^{u=1} = \dfrac{\log 2}2$$
A: Use instead
$$\int\frac1{x(1+x^2)}\mathrm dx=\log\varphi(x),\qquad\varphi(x)=\frac{x}{\sqrt{1+x^2}},
$$
and note that $\varphi(1)=\frac1{\sqrt2}$ and $\varphi(+\infty)=1$.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\overbrace{\color{#66f}{\large\int_{1}^{\infty}{\dd x \over x + x^{3}}}}
^{\ds{x \equiv \expo{t}}}\ =\
\int_{0}^{\infty}{\expo{-2t} \over 1 + \expo{-2t}}\,\dd t
=\left.{\ln\pars{1 + \expo{-2t}} \over -2}\,\right\vert_{\, 0}^{\,\infty}
=\color{#66f}{\large\half\,\ln\pars{2}}
\end{align}
A: We'll pick up where you wrote $$\int_1^\infty\frac{1}{x}-\frac{x}{(1+x^2)}dx.$$ Note that we have $$\lim_{N\to\infty}\left[\ln(x)-\frac{1}{2}\ln(1+x^2)\right]_{1}^N=\lim_{N\to\infty}\left[\ln(N)-\frac{1}{2}\ln(1+N^2)-\left(\ln(1)-\frac{1}{2}\ln(2)\right)\right].$$ Simplifying the natural logs, we have $$\lim_{N\to\infty}\left[\ln\left(\frac{N}{\sqrt{1+N^2}}\right)+\frac{\ln(2)}{2}\right].$$ It is easy to see that the argument of $\ln$ tends to $1$, thus the first term goes to $0$. This leaves us with the desired answer.
A: Letting $x=\sqrt u$, our integral becomes
$$\frac{1}{2}\int_1^{\infty}\frac{1}{u(u+1)} \mathrm{du}=\left[\frac{1}{2}\ln \left(\frac{u}{u+1}\right)\right]_1^{\infty}=\frac{\ln 2}{2}$$
Chris.
