proving an integral-defined function is differentiable? Im studying the following exercise on generalised Riemann integrals (I'm a self-learner): 
$$L(x)=\int_0^{+\infty}\frac{\ln(1+x^2t^2)}{1+t^2}dt$$
What I managed to do :  


*

*Prove that L is well-defined on $\mathbb{R}$ and even.

*Prove that $$\frac{2xt^2}{(1+x^2t^2)(1+t^2)}=\frac{2x}{1-x^2}\left(\frac{1}{1+x^2t^2}-\frac{1}{1+t^2}\right)$$


*using the result of question 3. (below) prove that $$\forall x \in \left]-1 ;1\right[, \  L(x)=\frac{\pi}{\ln(x+1)}$$
What I couldn't prove 


*Prove that L is differentiable on $\mathbb{R}\setminus\left\lbrace-1;1\right\rbrace \,$ and that $$L'(x)=\frac{\pi}{1+x}$$
I have trouble finding an upper bound to prove differentiability rigourously,,  according to the differentiation theorem, it's the absolute value of the patial derivative wrt to $x$ that need be bounded
Any help would be appreciated 
Thanks !
 A: Given $b>0$, we have the bound
$$
\frac{\ln(1+x^2t^2)}{1+t^2}\leq\frac{C}{(1+t^2)^{1-\varepsilon}} ,
\qquad
|x|\leq b,
$$
for any $\varepsilon>0$, with $C$ depending on $\varepsilon$ and $b$.
A: The final answers for $L(x)$ and $L'(x)$ mentioned in the question, are incorrect.
Note that for a given $x$, arbitrary and small $\alpha>0$ and large enough $t$, say $t>M(x,\alpha)$ we have $$\ln(1+x^2t^2)<(x^2t^2)^\alpha$$hence $$L(x)=\int_0^{+\infty}\frac{\ln(1+x^2t^2)}{1+t^2}dt{=\int_0^{M(x,\alpha)}\frac{\ln(1+x^2t^2)}{1+t^2}dt+\int_{M(x,\alpha)}^{+\infty}\frac{\ln(1+x^2t^2)}{1+t^2}dt\\\le\int_0^{M(x,\alpha)}\frac{\ln(1+x^2t^2)}{1+t^2}dt+\int_{M(x,\alpha)}^{+\infty}\frac{(x^2t^2)^\alpha}{1+t^2}dt\\\le\int_0^{M(x,\alpha)}\frac{\ln(1+x^2t^2)}{1+t^2}dt+\int_{0}^{+\infty}\frac{(x^2t^2)^\alpha}{1+t^2}dt}$$both the latter integrals exist and are bounded, hence $L(x)$ is well-defined over $\Bbb R\quad\blacksquare$ 
Based on the well-definition of $L(x)$, for $x\ne -1,0,1$ we can write$$L'(x){
={d\over dx}\int_0^{+\infty}\frac{\ln(1+x^2t^2)}{1+t^2}dt\\
=\int_0^{+\infty}{d\over dx}\frac{\ln(1+x^2t^2)}{1+t^2}dt\\
=\int_0^{+\infty}\frac{2xt^2}{(1+t^2)(1+x^2t^2)}dt\\
={2x\over x^2-1}\int_0^{+\infty}{1\over 1+t^2}-{1\over 1+x^2t^2}dt\\
={2x\over x^2-1}{\pi\over 2}\left[1-{1\over |x|}\right]\\
={\pi\over |x|+1}\times \text{sgn}(x)
}
$$Also simply $L'(0)=0$ and for $x=\pm 1$ we have $$L'(\pm 1){=\pm \int_0^\infty {2t^2\over (1+t^2)^2}dt\\=\pm 
\left[\tan^{-1}x-{x\over x^2+1}\right]_0^\infty
\\=\pm{\pi\over 2}
}$$finally

$$L'(x)={\pi\over |x|+1}\text{sgn}(x)\quad,\quad x\in\Bbb R$$

Since $L(0)=0$ we obtain 

$$L(x)=\pi \ln(1+|x|)$$

A: $$L′(x) =  \frac {π}{|x|+1sgn(x)}$$ 
so $ L'(x)\to \pi$ if $x \to 0^+$ and $L'(x) \to-\pi $ if $ x \to 0^-$, so $L'(0)$ can't be defined.
