Bt 1: Fill & Download for Free

This automatically destroys you answers. 0-0-0 droid and BT-1.BT-1 is armed to the teeth with weapons.0–0–0 is the ultimate interrogation droid. he literally “accidentally” killed someone while interrogating them.BT-1 with all extensions(ecxept the rocket launcher. theres supposed to be rocket launcher. and i WANT IT BACK)but(t) then again, HK-47,a jedi assassin droid, made by darth revan, might as well snipe them from a cliff and wipe them from the face of galaxy.

Botulinum Toxin. I have fibromyalgia. In the early 90s, physicians were doing test studies using the toxin, calling it BT-1. I used to joke and say I would be BaTman if I was able to get the treatment. Never happened, the research went to a different part of the medical community. Plastic/cosmetic surgery. BT-1 is injected with a needle, but a drop of it on your tongue could kill you instantly. BT-1 is botulism, a plague. BT-1 is also now called...Botox.

We want to evaluate the integral (where [math]a > 0[/math])[math]I(a) = \displaystyle \int_0^a \frac{\ln(x^2 + a^2)}{x + a} \, dx. \tag*{}[/math]To this end, we first let [math]x = at[/math]:[math]\begin{align*} I(a) &= \displaystyle \int_0^1 \frac{\ln(a^2(t^2 + 1))}{a(t + [/math][math]1[/math][math])} \cdot a \, dt\\ &= \displaystyle \int_0^1 \frac{2\ln{a} + \ln(t^2 + 1)}{t + [/math][math]1[/math][math]} \, dt\\ &= \displaystyle 2\ln{a} \ln(t + [/math][math]1[/math][math]) \Bigg|_0^1 + \int_0^1 \frac{\ln(t^2 + 1)}{t + [/math][math]1[/math][math]} \, dt\\ &= \displaystyle 2\ln{a} \ln{2} + \int_0^1 \frac{\ln(t^2 + 1)}{t + [/math][math]1[/math][math]} \, dt. \end{align*} \tag*{}[/math]Now, we need to evaluate[math]\displaystyle \int_0^1 \frac{\ln(t^2 + 1)}{t + [/math][math]1[/math][math]} \, dt. \tag*{}[/math]In order to do this, we start by using integration by parts with [math]u = \ln(t^2 + [/math][math]1[/math][math])[/math] and [math]dv = \frac{dt}{t + [/math][math]1[/math][math]}[/math]:[math]\begin{align*} \displaystyle \int_0^1 \frac{\ln(t^2 + 1)}{t + [/math][math]1[/math][math]} \, dt &= \ln(t^2 + [/math][math]1[/math][math]) \ln(t + [/math][math]1[/math][math]) \Bigg|_0^1 - 2\int_0^1 \frac{t \ln(t + 1)}{t^2 + [/math][math]1[/math][math]} \, dt\\ &= \displaystyle \ln^2{2} - 2\int_0^1 \frac{t \ln(t + 1)}{t^2 + [/math][math]1[/math][math]} \, dt. \end{align*} \tag*{}[/math]For the remaining integral, we (now) use a parameter and consider[math]\displaystyle J(b) = \int_0^1 \frac{t \ln(bt + 1)}{t^2 + [/math][math]1[/math][math]} \, dt. \tag*{}[/math]Note that we want the value of this integral when [math]b = [/math][math]1[/math][math][/math].Differentiating both sides with respect to [math]b[/math] yields[math]\begin{align*} J'(b) &= \displaystyle \int_0^1 \frac{\partial}{\partial b} \frac{t \ln(bt + 1)}{t^2 + [/math][math]1[/math][math]} \, dt\\ &= \displaystyle \int_0^1 \frac{t^2}{(bt + 1)(t^2 + [/math][math]1[/math][math])} \, dt\\ &= \displaystyle \frac{1}{b^2 + [/math][math]1[/math][math]} \int_0^1 \Big(\frac{1}{bt + [/math][math]1[/math][math]} + \frac{bt - 1}{t^2 + 1}\Big) \, dt, \text{ by partial fractions}\\ &= \displaystyle \frac{1}{b^2 + [/math][math]1[/math][math]} \Big(\frac{1}{b} \ln(bt + [/math][math]1[/math][math]) + \frac{b}{2} \ln(t^2 + [/math][math]1[/math][math]) - \arctan{t}\Big) \Bigg|_0^1\\ &= \displaystyle \frac{\ln(b + 1)}{b(b^2 + [/math][math]1[/math][math])} + \frac{\ln{2}}{2} \cdot \frac{b}{b^2 + [/math][math]1[/math][math]} - \frac{\pi}{4} \cdot \frac{1}{b^2 + [/math][math]1[/math][math]}. \end{align*} \tag*{}[/math]For the integral we want, now we integrate both sides from 0 to 1:[math]\begin{align*} J(1) &= \displaystyle \int_0^1 \Big(\frac{\ln(b + 1)}{b(b^2 + [/math][math]1[/math][math])} + \frac{\ln{2}}{2} \cdot \frac{b}{b^2 + [/math][math]1[/math][math]} - \frac{\pi}{4} \cdot \frac{1}{b^2 + 1}\Big) \, db\\ &= \displaystyle \int_0^1 \Big(\frac{\ln(b + 1)}{b} - \frac{b \ln(b + 1)}{b^2 + [/math][math]1[/math][math]} + \frac{\ln{2}}{2} \cdot \frac{b}{b^2 + [/math][math]1[/math][math]} - \frac{\pi}{4} \cdot \frac{1}{b^2 + 1}\Big) \, db\\ &= \displaystyle \int_0^1 \frac{\ln(b + 1)}{b} \, db - J(1) + \frac{\ln^2{2}}{4} - \frac{\pi^2}{16}. \end{align*} \tag*{}[/math]Solving for [math]J(1)[/math] gives us[math]J(1) = \displaystyle \frac{1}{2} \int_0^1 \frac{\ln(b + 1)}{b} \, db + \frac{\ln^2{2}}{8} - \frac{\pi^2}{32}. \tag*{}[/math]In order to evaluate the remaining integral, we use the Maclaurin series for [math]\ln(x+1)[/math] and obtain[math]\begin{align*} \displaystyle \int_0^1 \frac{\ln(b + 1)}{b} \, db &= \displaystyle \int_0^1 \frac{1}{b} \Big(\sum_{k=1}^{\infty} (-1)^{k-1} \frac{b^k}{k}\Big) \, db\\ &= \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \int_0^1 b^{k-1} \, db\\ &= \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2}\\ &= \displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} - \sum_{k=1}^{\infty} \frac{2}{(2k)^2}\\ &= \displaystyle \Big(1 - \frac{2}{2^2}\Big) \sum_{k=1}^{\infty} \frac{1}{k^2}\\ &= \displaystyle \frac{1}{2} \cdot \frac{\pi^2}{6}\\ &= \displaystyle \frac{\pi^2}{12}. \end{align*} \tag*{}[/math]Therefore, we have found that[math]J(1) = \displaystyle \int_0^1 \frac{t \ln(t + 1)}{t^2 + [/math][math]1[/math][math]} \, dt = \frac{1}{2} \cdot \frac{\pi^2}{12} + \frac{\ln^2{2}}{8} - \frac{\pi^2}{32} = \frac{\pi^2}{96} + \frac{\ln^2{2}}{8}. \tag*{}[/math]Putting this all together, we obtain[math]I(a) = \displaystyle 2 \ln{a} \ln{2} + \ln^2{2} - 2\Big( \frac{\pi^2}{96} + \frac{\ln^2{2}}{8} \Big) = 2 \ln{a} \ln{2} + \frac{3}{4} \ln^2{2} - \frac{\pi^2}{48}. \tag*{}[/math]In other words, we conclude that[math]\displaystyle \int_0^a \frac{\ln(x^2 + a^2)}{x + a} \, dx = 2 \ln{a} \ln{2} + \frac{3}{4} \ln^2{2} - \frac{\pi^2}{48}. \tag*{}[/math]