Setup
I'm taking a physics class and need to calculate (or approximate) a particular integral. Let $$\Delta(x):= 1-(1-x)x \frac{p^2}{m^2},$$ where $p^2:= p_\mu p_\nu\eta^{\mu\nu}$ with $\eta$ being the Minkowski metric. I'd like to show that approximately $$\int_0^1 dx\, \ln\left({\Delta(x)}\right)\propto \sqrt{1-\frac{4m^2}{p^2}}\ln\left(\frac{2m^2-p^2+\sqrt{p^2(p^2-m^2)}}{2m^2}\right).$$
My attempt
My only idea was to Taylor the $\ln$. Assuming $p^2/m^2$ to be small we have $$\ln(\Delta(x)) = \left(x^2-x\right)\frac{p^2}{m^2}+\frac{1}{2} (x-1) x \left(x-x^2\right)\left(\frac{p^2}{m^2}\right)^2 +O\left(\left({p^2}/{m^2}\right)^3\right).$$ Then $$\begin{align*} \int_0^1dx \ln\Delta(x) &\approx \int_0^1 dx\, \left[\left(x^2-x\right)\frac{p^2}{m^2}+\frac{1}{2} (x-1) x \left(x-x^2\right)\left(\frac{p^2}{m^2}\right)^2\right]\\ &= -\frac{1}{6}\frac{p^2}{m^2} - \frac{1}{60}\left(\frac{p^2}{m^2}\right)^2. \end{align*}$$ This seems completely off, or at least way too crude of an approximation.. Unfortunately, I don't really have much experience with integration so I don't really have any other smart ideas or tricks up my sleeve...
P.S. I have no idea how to name this question properly. If the title doesn't fit, please feel free to edit it.
You can easily have the exact result.
Let $$I=\int_0^1\log \big[1-k (1-x) x\big]\,dx$$ $$1-k (1-x) x=k x^2-k x+1=k(x-a)(x-b)$$ $$\log \big[1-k (1-x) x\big]=\log(k)+\log(x-a)+\log(x-b)$$ Using integration by parts $$J=\int \log(x-c)\,dx=(x-c) \log (x-c)-x+ C$$ $$K(c)=\int_0^1 \log(x-c)\,dx=-(c-1) \log (1-c)+c \log (-c)-1$$
Now, use the above for $$a=\frac{k+\sqrt{k-4} \sqrt{k}}{2 k} \qquad \text{and} \qquad b=\frac{k-\sqrt{k-4} \sqrt{k}}{2 k}$$
Edit
Rearranging all the logarithms leads to the simple result $$\int_0^1 \ln\left({\Delta(x)}\right)\,dx=\frac{2 \sqrt{4 m^2-p^2} }{p}\tan ^{-1}\left(\frac{p}{\sqrt{4 m^2-p^2}}\right)-2$$ which is exact and looks simpler then the looked for approximation.
Expanded as a infinite series, this gives $$\int_0^1 \ln\left({\Delta(x)}\right)\,dx=-\sum_{n=1}^\infty \frac{n!\,(n+1)!}{(2 n+3)!}\,\left(\frac{p^2}{m^2}\right)^n$$ In your post, you wrote the first and second terms.
This is quickly convergence since $$a_n=\frac{n!\,(n+1)!}{(2 n+3)!}\implies \frac{a_{n+1}}{a_n}=\frac{n+1}{4 n+10}$$