Before anything else, note that $f(0)=0$, so our constant of integration will be $0$.
As you found,
$$f'(a)=\int_0^\pi\frac{\cos x}{1+a\cos x}dx$$
$$f'(a)=\frac1a\int_0^\pi\frac{1-1+a\cos x}{1+a\cos x}dx$$
$$f'(a)=\frac1a\int_0^\pi\frac{1+a\cos x}{1+a\cos x}dx-\frac1a\int_0^\pi\frac{dx}{1+a\cos x}$$
$$f'(a)=\frac\pi a-\frac1aj(a)$$
For $j$, we preform $t=\tan\frac{x}2$ which gives
$$j(a)=2\int_0^\infty \frac{1}{1+a\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}$$
$$j(a)=\frac2{1-a}\int_0^\infty \frac{dt}{t^2+\frac{1+a}{1-a}}$$
It is easily shown that
$$\int_0^\infty \frac{dx}{x^2+c}=\frac\pi{2\sqrt c}$$
So we have that
$$j(a)=\frac\pi{(1-a)\sqrt{\frac{1+a}{1-a}}}=\frac\pi{\sqrt{1-a^2}}$$
Hence
$$f'(a)=\frac\pi a-\frac\pi{a\sqrt{1-a^2}}$$
So
$$f(a)=\pi\log|a|-\pi\int\frac{da}{a\sqrt{1-a^2}}$$
For the remaining integral, let $a=\sin x$ to get
$$\int\frac{da}{a\sqrt{1-a^2}}=\int\frac{dx}{\sin x}=-\log\left(\cot x+\csc x\right)=-\log\frac{1+\sqrt{1-a^2}}{a}$$
So
$$f(a)=\pi\log a+\pi\log\frac{1+\sqrt{1-a^2}}{a}$$
Which is just
$$f(a)=\pi\log\left(1+\sqrt{1-a^2}\right)$$
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\begin{align}
&\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\,
\cos\pars{x \over 2}\,\dd x \\[5mm] = &
\int_{0}^{\infty}{x\bracks{1 - 2\sin^{2}\pars{x/2}} - \sin\pars{x} \over x^{3}}\,\cos\pars{x \over 2}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{\infty}{2x\cos\pars{x/2} - 2x\sin\pars{x}\sin\pars{x/2} -
2\sin\pars{x}\cos\pars{x/2} \over x^{3}}\,\,\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{\infty}{x\cos\pars{x/2} + x\cos\pars{3x/2} -
\sin\pars{3x/2} - \sin\pars{x/2} \over x^{3}}\,\,\,\dd x
\\[1cm] & =
-\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -
{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & -
{1 \over 4}\int_{x\ =\ 0}^{x\ \to\ \infty}\bracks{2x - \sin\pars{3x/2} - \sin\pars{x/2}}\,\dd\pars{1 \over x^{2}}
\end{align}
Integrating by parts the last integral:
\begin{align}
&\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\,
\cos\pars{x \over 2}\,\dd x =
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -
{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & +
{1 \over 4}\int_{x = 0}^{\infty}{2 - 3\cos\pars{3x/2}/2 - \cos\pars{x/2}/2 \over x^{2}}\,\dd x
\\[1cm] & =
-\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -
{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & +
{3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x +
{1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x
\\[1cm] & =
-\,{3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x
-\,{1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & =
-\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x
-\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x =
-\,{3 \over 8}\
\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x
\\[5mm] & =
-\,{3 \over 4}\int_{0}^{\infty}{\sin^{2}\pars{x/2} \over x^{2}}\,\dd x =
-\,{3 \over 8}\
\underbrace{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x}
_{\ds{=\ {\pi \over 2}}}\ = \
\bbox[#ffe,10px,border:1px dotted navy]{\ds{-\,{3 \over 16}\,\pi}}
\end{align}
>By integrating by parts:
$\ds{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x =
\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x = {1 \over 2}\,\pi}$.
Best Answer
\begin{align} \int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x&\overset{\text{IBP}}=\Big[x\arctan(\sin x)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\arctan(\sin x)\,dx\\ &=\frac{\pi^2}{8}-\int_0^{\frac{\pi}{2}}\arctan(\sin x)\,dx\\ &\overset{u=\sqrt{\frac{1-\sin x}{1+\sin x}}}=\frac{\pi^2}{8}-2\int_0^1 \frac{\arctan\left(\frac{1-x^2}{1+x^2}\right)}{1+x^2}\,dx\\ &=\frac{\pi^2}{8}-2\int_0^1\frac{\arctan(1)}{1+x^2}\,dx+2\int_0^1\frac{\arctan(x^2)}{1+x^2}\,dx\\ &=2\int_0^1\frac{\arctan(x^2)}{1+x^2}\,dx\\ \int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ &=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \end{align}
Since,
$\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dt$
then,
\begin{align} \displaystyle K&=\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \displaystyle &=\int_0^1\int_0^1 \dfrac{2x^2}{(1+t^2x^2)(1+x^4)}dtdx\\ \displaystyle &=\int_0^1\int_0^1 \left(\dfrac{2t^2}{(1+t^4)(1+x^4)}+\dfrac{2x^2}{(1+x^4)(1+t^4)}-\dfrac{2t^2}{(1+t^4)(1+t^2x^2}\right)dtdx\\ &=\displaystyle 4\left(\int_0^1 \dfrac{t^2}{1+t^4}dt\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)-K \end{align}
Therefore,
$\displaystyle K=2\left(\int_0^1 \dfrac{x^2}{1+x^4}dx\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)$
Since,
\begin{align}\displaystyle \int_0^1 \dfrac{x^2}{1+x^4}dx&=\frac{1}{2\sqrt{2}}\left[\dfrac{1}{2}\ln\left(\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+\arctan\left(\sqrt{2}x+1\right)+\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi+\ln\left(3-2\sqrt{2}\right)\Big) \end{align}
and,
\begin{align}\displaystyle \int_0^1 \dfrac{1}{1+x^4}dx&=\dfrac{1}{2\sqrt{2}}\left[\dfrac{1}{2}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\arctan\left(\sqrt{2}x+1\right)+\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi-\ln\left(3-2\sqrt{2}\right)\Big) \end{align}
Therefore,
$\boxed{K=\displaystyle \dfrac{\pi^2}{16}-\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$
Since, $\displaystyle 3-2\sqrt{2}=\left(1-\sqrt{2}\right)^2=\frac{1}{\left(1+\sqrt{2}\right)^2}$
Therefore,
$\displaystyle \ln^2\left(3-2\sqrt{2}\right)=4\ln^2\left(1+\sqrt{2}\right)$
Thus,
$\boxed{\displaystyle\int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x=\dfrac{1}{2}\Big(\ln(1+\sqrt{2})\Big)^2}$