Simplifying $\int_0^t\frac{1}{1+Au^b}du$ I'm trying to simplify 
$$
\int_0^t\frac{1}{1+Au^b}du,\quad A>0,b>0,t\in[0,1].
$$ 
It looked simple at first but after trying a bit, I actually don't know how to tackle this. I entered the integral into Mathematica but it gave me back the same expression without any further simplification. Can someone help please?

A simple closed form in terms of elementary functions probably does not exist, but the integral can be expressed in terms of the Incomplete Beta Function, $\mathbf{B}\left(x;\alpha,\beta\right)$.
$$\int_{0}^{t}\frac{1}{1+Au^b}\operatorname{d}u = \frac{1}{bA^{\frac{1}{b}}}\mathbf{B}\left(\frac{At^b}{1+At^b};\frac{1}{b},1-\frac{1}{b}\right)$$
First make the substitution, $x = Au^b$. 
This brings the integral to the form
$$\frac{1}{bA^{\frac{1}{b}}}\int_{0}^{At^b}\frac{x^{\frac{1}{b}-1}}{1+x}\operatorname{d}x$$
which you can verify.
Now make the substitution $x = \frac{s}{1-s}$,
this brings the integral to the form
$$\frac{1}{bA^{\frac{1}{b}}}\int_{0}^{\frac{At^b}{1+At^b}}s^{\frac{1}{b}-1}(1-s)^{\left(1-\frac{1}{b}\right)-1}\operatorname{d}s = \frac{1}{bA^{\frac{1}{b}}}\mathbf{B}\left(\frac{At^b}{1+At^b};\frac{1}{b},1-\frac{1}{b}\right)$$
I'm not sure if this answer is useful to you, as this doesn't really simplify the integral. Perhaps it will serve as a good jumping off point.
