With integration of a geometric summation for \({1\over1-x}\), we get integrals of \(x^k\). If it was \({1\over1-x(1-x)}\), we would integrate \(x^k(1-x)^k\) which is the beta integral. Consider the following for positive values of \(c\). \[\int_0^1{dx\over1+cx(1-x)}=\sum_{k=0}^\infty(-1)^kc^k\int_0^1x^k(1-x)^kdx =\sum_{k=0}^\infty{(-1)^kc^k(k!)^2\over(2k+1)!}\] When \(0<c<4\), we get a convergent power series based on integrating beta functions. We can evaluate the integral to show that it gives us a way to create formulas for computing logarithms. \[\int_0^1{dx\over1+cx(1-x)}={1\over c}\int_0^1{dx\over{1\over c}+x-x^2} ={1\over c}\int_0^1{d\left(x-{1\over2}\right)\over\left({4+c\over4c}\right) -\left(x-{1\over2}\right)^2}\] Now let \(d=\sqrt{4+c\over4c}\) \[={1\over cd}\left[\text{arctanh} \left({x-{1\over2}\over d}\right)\right]_0^1 ={1\over2cd}\left[\log\left|{{1\over2}+d\over{1\over2}-d}\right| -\log\left|{-{1\over2}+d\over-{1\over2}-d}\right|\right]\] \[{1\over2cd}\log\left|{1+2d\over1-2d}\cdot{-1-2d\over-1+2d}\right| ={1\over2cd}\log\left|{(2d+1)^2\over(2d-1)^2}\right| ={1\over cd}\log\left|{1+2d\over1-2d}\right|\] Now if we were to want a formula for \(\log(r)\), we would need \[r={1+2d\over1-2d}\quad\text{or}\quad r=-{1+2d\over1-2d}\] So we obtain \[d={1\over2}\left({r+1\over r-1}\right)\quad\text{or}\quad d={1\over2}\left(r-1\over r+1\right)\] And solving for \(c\) from these, we get \[c={(r+1)^2\over-r}\quad\text{or}\quad c={(r-1)^2\over r}\] Considering that we only ever want \(r>0\), with a little algebra, we can show that we always get a smaller value of \(|c|\) by choosing \[c={(r-1)^2\over r}\] So this choice is better for faster series convergence. Next we substitute our results and obtain the following series for \(\log(r)\) \[\log(r)={r^2-1\over2r} \sum_{k=0}^\infty{(-1)^k\left({(r-1)^2\over r}\right)^k(k!)^2\over(2k+1)!}\] This series converges when \(|c|<4\) since \({(k!)^2\over(2k+1)!}\propto4^{-k}\), so \[\left|{(r-1)^2\over r}\right|<4\Rightarrow3-2\sqrt{2}<r<3+2\sqrt{2}\] Our series gives us the following explicit formulas for logs of integers \[\log(2)={3\over4}\sum_{k=0}^\infty{(-1)^k \left({1\over2}\right)^k(k!)^2\over(2k+1)!}\] \[\log(3)={4\over3}\sum_{k=0}^\infty{(-1)^k \left({4\over3}\right)^k(k!)^2\over(2k+1)!}\] \[\log(4)=2\log(2)={15\over8}\sum_{k=0}^\infty{(-1)^k \left({9\over4}\right)^k(k!)^2\over(2k+1)!}\] \[\log(5)={12\over5}\sum_{k=0}^\infty{(-1)^k \left({16\over5}\right)^k(k!)^2\over(2k+1)!}\] This series is not the most practical for computing individual logs, but if we were to compute a sequence of logs of prime integers, we could obtain \(\log(p_{n+1})\) from \(\log(p_n)\) by computing \[\log(p_{n+1})=\log(p_n)+\log\left({p_{n+1}\over p_n}\right)\] The ratio \(p_{n+1}/p_n\) would get close to \(1\) for most differences between primes which is good for convergence of our series. We would have \(r=1+\epsilon\) and an asymptotic ratio between terms of \({(r-1)^2\over4r}\approx\epsilon^2/4\).