I need help with this problem:
$$\int_0^1 \frac{\ln{(x+\sqrt{1-x^2})}}{x} \ dx = \frac{\pi^A}{B}$$ Find $A + B$.
I've tried all the analytical approaches I can think of, so I've been exploring series solutions but haven't come up with anything that works. My most promising attempt is as follows:
Substitute $x = \sin{(\theta)}$, $dx = \cos{(\theta)} \ d\theta$, thus the integral becomes
$$\int_0^{\frac{\pi}{2}}\frac{\ln{(\sin{\theta}+\cos{\theta})}}{\sin{\theta}}\cos{\theta} \ d\theta$$
Multiplying by $\dfrac{2}{2}$ leaves $\dfrac{1}{2}$ outside the integral and squares the inside of the natural log:
$$\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\ln{(1+2\sin{\theta}\cos{\theta})}}{\sin{\theta}}\cos{\theta} \ d\theta$$
I then used the series representation of $\ln{(1 + x)}$ and multiplied each term by $\dfrac{\cos{\theta}}{\sin{\theta}}$. I then integrated each term of the series from $n = 1$ to $n = 5$ (using Symbolab) but didn't get a result that lined up with any series representation for the form $\dfrac{\pi^A}{B}$
I know that there are a lot of series representations for the form $\dfrac{\pi^A}{B}$, which is why I'm using this approach.
I'm finishing calc 3 just for reference of my skill level. This was an honors project question that was assigned to me a while ago, and I never solved it, but it's been bothering me ever since. Can I get some help solving it?
