Why is $f(x)=\int_{1}^{x^2} \frac{\ln(xt)}{1+t}dt$ continuously differentiable and what is it's derivative?

In doing some old exam questions, I came across the following problem. Let $$f:(1,\infty)\rightarrow \mathbb{R}$$ $$f(x)=\int_{1}^{x^2} \frac{\ln(xt)}{1+t}dt$$ Questions:

a) Reason as to why f is continuously differentiable and find an integral-free representation of the derivative

b) Show that $$f'(x)>0$$ for $$x>1$$

I assume (b) will be trivial when (a) is solved.

For (a) my approach was integrating more or less directly, then differentiating. I end up caught in a cycle of integrating by parts: $$\int_{1}^{x^2} \frac{\ln(1+t)}{t}dt$$and $$\int_{1}^{x^2} \frac{\ln(t)}{t+1}dt$$ keep coming back... I don't have anything in my toolbox (that I know of) that lets me crack this.

So I assume that either I shouldn't actually integrate and then differentiate or I'm missing something in my "integration-toolbelt". What's going on here? Also, I don't actually know how to "Reason as to why f is continuously differentiable", are my problems connected?

Hint: Use the fact that $$\ln(xt)=\ln x+\ln t$$. So $$f(x)=\ln \, x\int_1^{x^{2}} \frac 1 {1+t} dt+\int_1^{x^{2}} \frac {\ln t } {1+t}dt$$ and the first term is $$(\ln x) (\ln (1+x^{2})-\ln 2)$$. You can now write down $$f'$$ easily.
It is also quite easy to show that $$f'(x) >0$$ for $$x >1$$. I will leave that to you.
$$let\; y(x)\;=\;\int_{1}^{x^2} \frac{\ln(xt)}{t+1}dt$$ Therefore y(x) = $$\int_{1}^{x^2} \frac{\ln(x)}{t+1}dt\;$$ + $$\int_{1}^{x^2} \frac{\ln(t)}{t+1}dt$$ Hence y(x) = $$\int_{1}^{x^2} \frac{\ln(t)}{t+1}dt\;$$ + $$ln(x)\, \int_{1}^{x^2} \frac{1}{t+1}dt$$  Now differentiate $$\frac{dy}{dx}\;= \frac{2xlnx^2}{1+x^2}\; + \; lnx\,\frac{2x}{1+x^2}\;+\; (\frac{1}{x})(\; \int_{1}^{x^2}\frac{1}{1+t})dt$$  since all terms are positive for x $$\gt1\;$$hence f'(x)$$\gt0$$