Computing an awful integral We can find (and Wolfram Alpha does it perfectly) that:
$$\pi e=\int_0^1\left(\frac{4 e^x \cos(\ln(x))}{x^2+1}-\frac{4 e^x \arctan(x) \sin(\ln(x))}x+4 e^x \arctan(x) \cos(\ln(x))\right)\mathrm d x.$$
We could prove that by noticing that 
$$\frac {\mathrm d}{\mathrm d x}\left(4\arctan(x)e^x\cos(\ln(x))\right)=\frac{4 e^x \cos(\ln(x))}{x^2+1}-\frac{4 e^x \arctan(x) \sin(\ln(x))}x+4 e^x \arctan(x) \cos(\ln(x))$$
but it seems like cheating.
Would there be a more systematic way to deal with such an integral ?
 A: You could start by finding $\displaystyle\int\frac{4e^x\cos(\ln x)}{x^2+1}dx\;$ using integration by parts 
with $\displaystyle u=4e^x\cos(\ln x),\;dv=\frac{1}{x^2+1}dx,\;du=\left(4e^x\cos(\ln x)-\frac{4e^x\sin(\ln x)}{x}\right)dx,\;v=\tan^{-1}x$ to get
$\displaystyle\int\left(\frac{4 e^x \cos(\ln(x))}{x^2+1}-\frac{4 e^x \arctan(x) \sin(\ln(x))}x+4 e^x \arctan(x) \cos(\ln(x))\right)dx$
$\displaystyle=\left[4e^x\cos(\ln x)\tan^{-1}x-\int\left(4e^x\cos(\ln x)\tan^{-1}x-\frac{4e^x\sin(\ln x)\tan^{-1}x}{x}\right)dx\right]$
$\displaystyle\hspace{1.5 in}+\int\left((4e^x\cos(\ln x)\tan^{-1}x-\frac{4e^x\sin(\ln x)\tan^{-1}x}{x}\right)dx$
$\displaystyle\hspace{.1 in}=\color{blue}{4e^x\cos(\ln x)\tan^{-1}x+C}$
A: How to find the solution naturally without cheating! (I mean without
previously knowing the solution). 
First, note that the given integral 
$$\color{blue}{
\begin{equation*}
I=\int \left( \frac{4e^{x}\cos \left( \ln x\right) }{1+x^{2}}-\frac{ 
4e^{x}\arctan \left( x\right) \sin \left( \ln x\right) }{x}+4e^{x}\arctan
(x)\ln (\cos (x))\right) dx
\end{equation*} }$$ 
can be re-written as
$$\color{blue}{ 
\begin{equation*}
\int \left( \frac{4\cos \left( \ln x\right) }{1+x^{2}}-\frac{4\arctan \left(
x\right) \sin \left( \ln x\right) }{x}+4\arctan (x)\ln (\cos (x))\right)
e^{x}dx
\end{equation*}}$$ 
which is the form of
$$\color{blue}{ 
\begin{equation*}
\int h(x)e^{x}dx.
\end{equation*}}$$ 
So, we recognize the well-known formula
$$\color{blue}{ 
\begin{equation*}
\int \left( f^{\prime }(x)+f(x)\right) e^{x}dx=f(x)e^{x}+C.
\end{equation*}}$$ 
The easy proof (based on integration by parts) for a more general formula can be found at
Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$
So the problem now is (without doing integral calculus) how to find $\color{blue}{f(x)}$
such that 
$$\color{blue}{
\begin{equation*}
4\left( \frac{\cos \left( \ln x\right) }{1+x^{2}}-\frac{\arctan \left(
x\right) \sin \left( \ln x\right) }{x}+\arctan (x)\ln (\cos (x))\right)
=f^{\prime }(x)+f(x).
\end{equation*}}$$
In the parentheses there are some well known derivatives expressions as 
$$\color{blue}{
\begin{eqnarray*}
\frac{1}{1+x^{2}} &=&\left( \arctan x\right) ^{\prime } \\
&& \\
\frac{1}{x} &=&\left( \ln x\right) ^{\prime }
\end{eqnarray*}}$$
so we can write 
$$\color{blue}{
\begin{equation*}
\cos \left( \ln x\right) \left( \arctan x\right) ^{\prime }-\sin \left( \ln
x\right) \left( \ln x\right) ^{\prime }\arctan \left( x\right) +\arctan
(x)\ln (\cos (x))
\end{equation*}}$$ 
with a little more experience (about chain rule) we can also note that 
$$\color{blue}{
\begin{equation*}
-\sin \left( \ln x\right) \left( \ln x\right) ^{\prime }=\left( \cos (\ln
x)\right) ^{\prime }
\end{equation*}}$$ 
so we have 
$$\color{blue}{
\begin{equation*}
\cos \left( \ln x\right) \left( \arctan x\right) ^{\prime }+\left( \cos (\ln x)\right) ^{\prime }\arctan \left( x\right) +\arctan (x)\ln (\cos (x))
\end{equation*}}$$ 
here we recognize the derivative of the product of two functions
$$\color{blue}{ 
\begin{equation*}
u(x)v^{\prime }(x)+u^{\prime }(x)v(x)=\left( u(x)v(x)\right) ^{\prime }
\end{equation*}}$$ 
where 
$$\color{blue}{
\begin{equation*}
u(x)=\ln (\cos x),\ \ \ \ \ \ \ \ \ \ \text{and} {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ }v(x)=\arctan x.
\end{equation*}}$$ 
So, we arrived to
$$\color{blue}{
\begin{equation*}
\left( \arctan x\ln (\cos x)\right) ^{\prime }+\arctan (x)\ln (\cos (x))
\end{equation*}}$$ 
which gives (naturally)
$$\color{blue}{
\begin{equation*}
f(x)=4\arctan (x)\ln (\cos (x))
\end{equation*}}$$
and therefore 
$$\color{blue}{
\begin{equation*}
I=\int (f^{\prime }(x)+f(x))e^{x}=4\arctan (x)\ln (\cos (x))e^{x}+C. 
\color{red} \blacksquare 
\end{equation*} }$$
