Show that $\lim_{n\rightarrow\infty}\int_{0}^{n}\frac{\sqrt x\ln x}{(1+x)^2}\,dx=\pi$ Show that $\lim_{n\rightarrow\infty}\int_{0}^{n}\frac{\sqrt x\ln x}{(1+x)^2}\,dx=\pi$. This limit puzzled me as I never worked on this kind before and I am inclined to think that, as the $n$ is in the integral's boundary, there is a theorem involved like the dominant convergence theorem. I tried integrating by parts, but the $0$ of the integral prevented me from applying $log0$. How should I handle it?
 A: The integrand is positive for $x>1$, so normally we'd just state the problem as $$\int_0^\infty\frac{\sqrt{x}\ln x}{(1+x)^2}dx=\pi.$$Let's first note that the substitution $x=\tan^2 t$ allows us to solve a seemingly unrelated problem, $$\int_0^\infty\frac{x^{k-1}}{(1+x)^2}dx=\int_0^{\pi/2}2\sin^{2k-1}t\cos^{3-2k} tdt=\operatorname{B}(k,\,2-k)\\=\Gamma(k)\Gamma(1-k)=\pi(1-k)\csc\pi k.$$(Look up Beta and Gamma functions if you don't know them well.) But this problem is not unrelated! Let's differentiate with respect to $k$: $$\int_0^\infty\frac{x^{k-1}\ln x}{(1+x)^2}dx=-\pi\csc\pi k[1+(1-k)\cot\pi k].$$Finally, substituting $k=\frac{3}{2}$ gives $$\int_0^\infty\frac{\sqrt{x}\ln x}{(1+x)^2}dx=\pi.$$
A: \begin{align}J&=\int_0^\infty \frac{\sqrt{x}\ln x}{(1+x)^2}\,dx\end{align}
Perform the change of variable $y=\sqrt{x}$
\begin{align}J&=\int_0^\infty \frac{4x^2\ln x}{(1+x^2)^2}\,dx\\
&=\left[-\frac{1}{2(1+x^2)}\times 4x\ln x\right]_0^\infty+\int_0^\infty \frac{2\ln x}{1+x^2}\,dx+\int_0^\infty \frac{2}{1+x^2}\,dx\\
&=\int_0^\infty \frac{2}{1+x^2}\,dx\\
&=2\Big[\arctan x\Big]_0^\infty\\
&=\pi
\end{align}
NB:
\begin{align}\int_0^\infty \frac{\ln x}{1+x^2}\,dx=0\end{align}
(perform the change of variable $y=\frac{1}{x}$ )
A: Hint:
You can use the substitution $\sqrt x=t\iff x=t^2,\;t>0$ and obtain for the indefinite integral:
$$\int \frac{\sqrt x\ln x}{(1+x)^2}\,dx=\int\frac{2t\ln t}{(1+t^2)^2}\,\frac{\mathrm dt}{t}=2\int\frac{\ln t}{(1+t^2)^2}\,\mathrm dt.$$
Now you get rid of the log with an integration by parts, setting
$$u=\ln t,\quad\mathrm dv=\frac{\mathrm d t}{(1+t^2)^2},\enspace\text{whence}\quad\mathrm d u=\frac{\mathrm dt}{t},\quad v= \int\frac{\mathrm d t}{(1+t^2)^2}.$$
This last integral is classically obtained from 
$$\arctan t=\int\frac{\mathrm d t}{1+t^2}$$
integrating the latter by parts (again!).
A: Let $x=\tan^2\theta$ and then
\begin{eqnarray*}
I&=&\int_0^\infty{\sqrt x\ln x\over(1+x)^2}\,dx\\
&=&\int_0^{\frac\pi2}\frac{2\tan\theta\ln(\tan\theta)}{\sec^4(\theta)}2\tan\theta\sec^2\theta\,d\theta\\
&=&4\int_0^{\frac\pi2}\sin^2\theta\ln(\tan\theta)\,d\theta\\
&=&2\int_0^{\frac\pi2}[1-\cos(2\theta)]\ln(\tan\theta)\,d\theta\\
&=&-2\int_0^{\frac\pi2}\cos(2\theta)\ln(\tan\theta)\,d\theta.
\end{eqnarray*}
Here
$$ \int_0^{\frac\pi2}\ln(\tan\theta)\,d\theta=0 $$
is used. Integration by parts gives
$$ I=-2\int_0^{\frac\pi2}\cos(2\theta)\ln(\tan\theta)\,d\theta=-2(-\theta+\frac12\sin(2\theta)\ln(\tan\theta))\bigg|_0^{\frac\pi2}=\pi. $$
A: Note
\begin{align}
\int_0^\infty \frac{\sqrt{x}\ln x}{(1+x)^2}\,dx
\overset{x\to 1/x}=& \>\>\frac12 \int_0^\infty \frac{(\sqrt{x}-\frac1{\sqrt x})\ln x}{(1+x)^2}\,dx\\
=&\>- \int_0^\infty \ln x \>d\left(\frac{\sqrt{x}}{1+x}\right)
\overset{ibp}=\int_0^\infty \frac{1}{\sqrt x(1+x)}dx
=\pi\
\end{align}
