Problems

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Solution:

1. \begin{align*} \sqrt{x^2+1} &= |x|\sqrt{1+\frac{1}{x^2}} \\ &=|x| \left (1 + \frac12 \frac{1}{x^2} + \cdots \right) & \text{if } \left (\frac{1}{x^2} < 1 \right) \\ &= |x| + \frac12 \frac{1}{|x|} + \cdots \\ &\approx |x| + \frac{1}{2|x|} \end{align*}
   

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2. 1. \begin{align*} && \tan( g(x) + g(-x)) &= \tan \left ( \tan^{-1}(\sqrt{x^2+1}-x) + \tan^{-1}(\sqrt{x^2+1}+x) \right) \\ &&&= \frac{\sqrt{x^2+1}-x+\sqrt{x^2+1}+x}{1-1} \\ \Rightarrow && g(x) + g(-x) &\in \left \{\cdots, -\frac{\pi}{2}, \frac{\pi}{2}, \cdots \right\} \end{align*} But \(g(x), g(-x) > 0\) and \(g(x), g(-x) \in (-\frac{\pi}{2}, \frac{\pi}{2})\), therefore it must be \(\frac{\pi}{2}\).
   2. \begin{align*} && \tan(2(k(x) + k(-x))) &= \tan(\tan^{-1}x + \tan^{-1}(-x)) \\ &&&= 0 \\ \Rightarrow && k(x)+k(-x) &\in \left \{\cdots, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \cdots \right\} \\ \end{align*} But \(k(x) \in (-\frac{\pi}{4}, \frac{\pi}{4})\), therefore \(k(x) + k(-x) = 0\).
   3. Let \(t = \tan k(x)\). \begin{align*} && \tan \left ( \tan^{-1} \frac{1}{x} \right) &= \frac{2 \tan\left ( \frac12 \tan^{-1} \frac1x \right)}{ 1-\tan^2\left ( \frac12 \tan^{-1} \frac1x \right)} \\ \Rightarrow && \frac1x &= \frac{2t}{1-t^2} \\ \Rightarrow && 1-t^2 &= 2tx \\ \Rightarrow && 0 &= t^2+2tx - 1 \\ \Rightarrow && 0 &= (t+x)^2 - 1-x^2 \\ \Rightarrow && t &= -x \pm \sqrt{1+x^2} \end{align*} Since \(t > 0\), \(t = \sqrt{1+x^2}-x = f(x) = \tan g(x)\)
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   5. \begin{align*} \int_0^1 k(x) \d x &= \int_0^1 \frac12 \tan^{-1} \left ( \frac1x \right) \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 - \int_0^1 \frac{x}{2} \frac{-1/x^2}{1+1/x^2} \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 + \frac14 \int_0^1 \frac{2x}{1+x^2} \d x \\ &= \frac12 \frac{\pi}{4} + \frac14 \ln(2) \\ &= \frac{\pi + \ln 4}{8}\end{align*} Therefore \(\displaystyle \int_{-1}^0 g(x) \d x = -\frac{\pi + \ln 4}{8}\)

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Solution:

1. \(\,\) \begin{align*} && f_\alpha(x) &= \tan^{-1}\!\left(\frac{x\tan\alpha + 1}{\tan\alpha - x}\right) \\ && f'_\alpha(x) &= \frac{1}{1 + \left(\frac{x\tan\alpha + 1}{\tan\alpha - x}\right) ^2} \cdot \frac{\tan \alpha \cdot (\tan \alpha - x) - (x \tan \alpha + 1) \cdot (-1)}{(\tan \alpha - x)^2} \\ &&&= \frac{\tan^2 \alpha -1}{(\tan \alpha - x)^2 + (x \tan \alpha +1)^2} \\ &&&= \frac{\tan^2 \alpha +1}{\tan^2 \alpha - 2x \tan \alpha + x^2 + x^2 \tan^2 \alpha + 2 x \tan \alpha + 1} \\ &&&= \frac{1+\tan^2 \alpha}{(1+\tan^2 \alpha(x^2 + 1)} = \frac{1}{1+x^2} \end{align*}
   

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2. Let \(g(x) = \tanh^{-1}(\sin x) - \sinh^{-1}(\tan x)\) then \begin{align*} && g'(x) &= \frac{1}{1-\sin^2 x} \cdot \cos x - \frac{1}{\sqrt{\tan^2 +1}} \cdot \sec^2 x \\ &&&= \sec x - \frac{\sec^2 x}{|\sec x|} \\ &&& = \begin{cases} 0 &\text{if } \sec x \geq 0 \\ 2 \sec x &\text{ otherwise} \end{cases} \end{align*} Therefore \(g'(x) = 2\sec x\) if \(\tfrac12 \pi < x < \tfrac32\pi\) Therefore $\displaystyle g(x) = \begin{cases} 0 & \text{if } x \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \\ \ln( (\tan x + \sec x)^2) + C &\text{otherwise} \end{cases}$
   

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Solution:

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2. \(\,\) \begin{align*} && y &= \frac{cx}{\sqrt{x^2+p}} \\ && \d y &= \frac{c(x^2+p)-cx^2}{(x^2+p)^{3/2}} \d x \\ &&&= \frac{cp^2}{(x^2+p)^{3/2}} \d x\\ && I &= \int \frac1{(b^2-y^2)\sqrt{c^2-y^2}} \d y \\ &&&= \int \frac{1}{\left ( b^2 - \frac{c^2x^2}{x^2+p} \right) \sqrt{c^2 - \frac{c^2x^2}{x^2+p} }} \d y \\ &&&= \int \frac{(x^2+p)^{3/2}}{((b^2-c^2)x^2+pb^2)\sqrt{c^2p}}\frac{cp}{(x^2+p)^{3/2}} \d x \\ &&&= \int \frac{\sqrt{p}}{((b^2-c^2)x^2+pb^2)} \d x \\ p=1: &&&= \int \frac{1}{(b^2-c^2)x^2+b^2} \d x \end{align*} When \(b = \sqrt{3}, c = \sqrt{2}\) \begin{align*} && I_1 &= \int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \d y\\ &&&= \int_{x =1 }^{x=\infty} \frac{1}{3+x^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} \right]_1^\infty \\ &&&= \frac{\pi}{2\sqrt{3}} - \frac{1}{\sqrt{3}} \frac{\pi}{6} \\ &&&= \frac{\pi}{3\sqrt{3}} \end{align*} \begin{align*} && I_2 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = \frac1y, \d x = -\frac1{y^2} \d y &&&= \int_{x=\sqrt{2}}^{x=1} \frac{x^2}{(3-x^2)\sqrt{2-x^2}}\cdot \left ( -\frac{1}{x^2} \right ) \d x \\ &&&= \int_1^{\sqrt{2}} \frac{1}{(3-x^2)\sqrt{2-x^2}} \d x \\ &&&= I_1 = \frac{\pi}{3\sqrt{3}} \end{align*}
3. \(\,\) \begin{align*} && I_3 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = 1/y, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{2}} \frac{x}{(3-x^2)\sqrt{2-x^2}} \d x \\ u = x^2, \d u = 2x \d x &&&= \int_{u=1}^{u=2} \frac{\frac12}{(3-u)\sqrt{2-u}} \d u \\ v=2-u, \d v = -\d u &&&= \frac12\int_{v=0}^{v=1} \frac{1}{(1+v)\sqrt{v}} \d v \\ &&&=\left [\tan^{-1}\sqrt{v}\right]_0^1 \\ &&&= \frac{\pi}{4} \end{align*}

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Solution:

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2. \(\,\)
   

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3. \(\,\)
   

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4. Note that \(\frac{x+x^4}{1+x^2+x^4}\) is continuous, and nicely behaved on \((-\infty, \infty)\) so we can see that \(\lim_{x \to \infty} h_1(x) = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\). \(\frac{4x-x^3}{1-x^4}\) on the other hand has asymptotes at \(\pm 1\). So as as \(x \to 1\), \(h_1(x) \to \pi + \frac{\pi}{2} = \frac{3\pi}{2}\). Then as \(x \to \infty\) we increase by another \(\frac{\pi}{2}\), so \(\lim_{x \to \infty} h_2(x) = 2\pi\)

An alternative way to think about the last two parts is to consider \(h\) as giving the (continuous) argument (shifted by \(\pi\)) of \((1-t^2)+it\) (blue), \((1+t^2+t^4)+i(t+t^4)\) (orange) or \((1-t^4)+i(4t-t^3)\) (green). We can see the orange line never wraps around the origin, so the argument is always easy to find. The blue does one full circuit, from \(-\pi\) to \(\pi\) (or \(0\) to \(2\pi\) in our world. And the green line also does a full \(2\pi\) loop.

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Solution:

1. \begin{align*} && \int \frac{\sinh x}{\cosh 2x} \d x &= \int \frac{\sinh x}{2\cosh^2 x -1} \d x \\ u = \cosh x, \d u = \sinh x \d x &&&= \int \frac{1}{2u^2 -1} \d u \\ &&&= \int\frac12 \left ( \frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1} \right) \d u \\ &&&= \frac1{2\sqrt{2}} \left (\ln (\sqrt{2}u-1) - \ln(\sqrt{2}u+1) \right) + C \\ &&&= \frac{1}{2\sqrt{2}} \ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) + C \end{align*}
2. \begin{align*} && \int \frac{\cosh x}{\cosh 2x} \d x &= \int \frac{\cosh x}{1+2\sinh^2 x} \d x \\ u = \sinh x && &= \int \frac{1}{1+2u^2} \d u \\ &&&=\frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}u) + C \\ &&&= \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) + C \end{align*}
3. \begin{align*} u = e^x : && \int_0^1 \frac{1}{1+u^4} \d u &= \int_{x=-\infty}^{x=0} \frac{1}{1+e^{4x}}e^{x} \d x \\ &&&= \int_{-\infty}^{0} \frac{e^{-x}}{e^{2x}+e^{-2x}} \d x \\ &&&= \int_{-\infty}^{0} \frac{\cosh x - \sinh x}{2\cosh 2x } \d x \\ &&&= \frac12 \int_{-\infty}^{0} \frac{\cosh x}{\cosh 2x} \d x - \frac12 \int_{-\infty}^{0} \frac{\sinh x}{\cosh 2x} \\ &&&= \frac12 \left [\frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) \right]_{-\infty}^{0}-\frac12 \left [ \frac{1}{2\sqrt{2}}\ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) \right]_{-\infty}^{0} \\ &&&= 0 - \frac1{2\sqrt{2}} \frac{-\pi}{2}-\left (\frac1{4\sqrt{2}} \ln \left (\frac{\sqrt{2}-1}{\sqrt{2}+1} \right) - 0 \right) \\ &&&= \frac{\pi - \ln((\sqrt{2}-1)^2)}{4\sqrt{2}} \\ &&&= \frac{\pi + 2 \ln(1+\sqrt{2})}{4\sqrt{2}} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && f(x) &= \frac{x+\sqrt3}{1-\sqrt3x} \\ \Rightarrow && f(f(x)) &= \frac{f(x)+\sqrt3}{1-\sqrt3f(x)} \\ &&&= \frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{x+\sqrt{3}+\sqrt3(1-\sqrt3x)}{1-\sqrt3x-\sqrt3(x+\sqrt3)} \\ &&&= \frac{-2x+2\sqrt3}{-2-2\sqrt3x} \\ &&&= \frac{x-\sqrt3}{1+\sqrt3 x} \\ \\ && f^3(x) &= f^2(f(x)) \\ &&&= \frac{f(x)-\sqrt3}{1+\sqrt3 f(x)} \\ &&&=\frac{\frac{x+\sqrt3}{1-\sqrt3x}-\sqrt3}{1+\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{(x+\sqrt3)-\sqrt3(1-\sqrt3 x)}{(1-\sqrt3x)+\sqrt3 (x+\sqrt3)} \\ &&&= \frac{-2x}{-2} = x \\ \\ && f^{2007}(x) &= x \end{align*}
2. If \(x = \tan \theta\) then \(f(x) = \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \tan \theta} = \tan (\theta + \frac{\pi}{3})\) and hence \(f^n(x) = \tan (\theta + \frac{n \pi}{3})\)
3. Note that if \(t = \sin \theta\) then \(g(t) = \cos \frac{\pi}{6} t\sin \theta + \frac12 \cos \theta = \sin(\theta + \frac{\pi}6)\) therefore \(g^n(t) = \sin(\sin^{-1}(t) + \frac{n\pi}{6})\)

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Solution:

1. \begin{align*} && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ u = v^{-1}, \d u = -v^{-2} \d v&&&= \int_{v = \infty}^{v = 1/x} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= \int_{1/x}^\infty \frac{1}{1+v^2} \d v \\ \\ \Rightarrow && t(x) + t(1/x) &= \int_0^x \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_{1/x}^{\infty} \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u \\ &&&= \frac12 p \\ \\ \Rightarrow && t(1) +t(1/1) = 2t(1) &= \frac12 p \end{align*}
2. \(\,\) \begin{align*} && y &= \frac{u}{\sqrt{1+u^2}} \\ \Rightarrow && y^2 &= \frac{u^2}{1+u^2} \\ &&&= 1-\frac{1}{1+u^2} \\ \Rightarrow && 1+u^2 &= \frac{1}{1-y^2} \\ \Rightarrow && u &= \frac{y}{\sqrt{1-y^2}} \\ \\ && \frac{\d u}{\d y} &= \frac{\sqrt{1-y^2} + y^2(1-y^2)^{-1/2}}{1-y^2} \\ &&&= \frac{1}{(1-y^2)^{3/2}} \\ \\ && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_0^{y = x/\sqrt{1+x^2}} \frac{1}{1 + \frac{y^2}{1-y^2}} \frac{1}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1-y^2}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1}{(1-y^2)^{1/2}} \d y \\ &&&= s\left ( \frac{x}{\sqrt{1+x^2}} \right) \\ \\ \Rightarrow && s\left ( \frac{1}{\sqrt{2}} \right) &= t(1) = \frac14p \end{align*}
3. \(\,\) \begin{align*} && z &= \frac{u + \frac1{\sqrt{3}}}{1- \frac{1}{\sqrt{3}} u}\\ \Rightarrow && z - \frac{z}{\sqrt{3}}u &= u + \frac{1}{\sqrt{3}} \\ \Rightarrow && u &= \frac{z-\frac{1}{\sqrt{3}}}{1 + \frac{z}{\sqrt{3}}} \\ \\ \Rightarrow && \frac{\d u}{\d z} &= \frac{\sqrt{3}(\sqrt{3}+z ) -(\sqrt{3}z-1)}{\left (\sqrt{3}+z \right)^2} \\ &&&= \frac{4}{(\sqrt{3}+z)^2} \\ \\ \Rightarrow && t \left ( \frac{1}{\sqrt{3}} \right) &= \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \d u \\ &&&= \int_{z=1/\sqrt{3}}^{z=\sqrt{3}} \frac{1}{1 + \left ( \frac{\sqrt{3}z-1}{\sqrt{3}+z}\right)^2} \frac{4}{(\sqrt{3}+z)^2} \d z\\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{(\sqrt{3}+z)^2+(\sqrt{3}z-1)^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{4+4z^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \d z \end{align*} Notice that \(t(1/\sqrt{3})+t(\sqrt{3}) = \frac12p\) and also notice that \(t(1/\sqrt{3}) + t(1/\sqrt{3}) =t(\sqrt{3})\) so \(3t(1/\sqrt{3}) = \frac12p\)

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Solution: \begin{align*} &&2f(x) &\equiv f(x) + f(x) \\ &&&\equiv f(x+x) \\ &&&\equiv f(2x) \\ \\ \Rightarrow && 2f(0) &= f(0) \\ \Rightarrow && f(0) &= 0 \\ && f''(0) &= \lim_{h \to 0} \frac{f(2h)-2f(0)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(2h)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(0)}{h^2} \\ &&&= 0 \\ \Rightarrow && f''(0) &= 0 \end{align*} If \(f(x)\) satisfies the equation, then \(f'(x)\) satisfies the equation. In particular this means that \(f^{(n)}(0) = 0\) for all \(n \geq 2\). Therefore the only non-zero term in the Maclaurin series is \(x^1\). Therefore \(f(x) = cx\)

1. Suppose \(g(x)g(y) \equiv g(x+y)\), then if \(G(x) = \ln g(x)\) we must have \(G(x)+G(y) \equiv G(x+y)\), ie \(G(x) = cx \Rightarrow g(x) = e^{cx}\)
2. Suppose \(h(x)+h(y) \equiv h(xy)\), then if \(h(e^u) = H(u)\) we must have that \(H(u)+H(v) \equiv h(e^u) + h(e^v) \equiv h(e^{u+v}) \equiv H(u+v)\).Therefore \(H(u) = cu\), ie \(h(e^u) = cu\) or \(h(x) = h(e^{\ln x}) = c \ln x\).
3. Finally if \(t(x) + t(y) \equiv t(z)\), the considering \(T(w) = t(\tan w)\) then \(T(x) + T(y) \equiv t(\tan x) + t(\tan y) \equiv t( \frac{\tan x + \tan y}{1- \tan x \tan y}) \equiv t (\tan (x+y)) \equiv T(x+y)\). Therefore \(T(x) = cx\) Therefore \(t(\tan w) = c w \Rightarrow t(x) = c \tan^{-1} x\)

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Solution: \begin{align*} && \tan \left (\arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) \right) &= \frac{\frac1{a+b}+\frac1{a+c}}{1-\frac{1}{(a+b)(a+c)}} \\ &&&= \frac{a+c+a+b}{(a+b)(a+c)-1} \\ &&&= \frac{2a+b+c}{a^2+ab+ac+bc-1} \\ &&&= \frac{2a+b+c}{2a^2+ab+ac} \\ &&&= \frac{1}{a} \\ &&&= \tan \arctan \frac1a\\ \Rightarrow && \arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) &= \arctan \frac{1}{a} + n \pi \end{align*} Since \(\arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2})\) the LHS \(\in (0, \pi)\) so \(n = 0\). \begin{align*} a=p+q, b = s, c = t:&& \arctan \! \!\left(\!\frac1 {p+q+s}\!\right) + \arctan \! \!\left(\!\frac 1{p+q+t}\!\right) &= \arctan \left ( \frac{1}{p+q} \right) \\ a=p+r, b= u, c = v && \arctan \! \!\left(\!\frac 1 {p+r+u}\!\right) + \arctan \! \!\left(\!\frac1 {p+r+v}\!\right) &= \arctan \! \!\left(\!\frac1 {p+r}\!\right) \\ a = p, b = q, c = r:&& \arctan \left ( \frac{1}{p+q} \right) +\arctan \! \!\left(\!\frac1 {p+r}\!\right) &= \arctan \left ( \frac1p \right) \end{align*} and the result follows. Taking \(p = 7\) we need to solve \[ \begin{cases} q+s &= 6 \\ q+t &= 14 \\ r+u &= 75 \\ r+v &= 180 \end{cases} \] also satisfying \(qr = 50\) etc, so say \(q = 1, r = 50, s = 5, v=25\)

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Solution: \begin{align*} u = x^m, \d u = m x^{m-1} && \int \frac{m}{x f(x^m)} \d x &= \int \frac{m x^{m-1}}{uf(u)} \d x \\ &&&= \int \frac{1}{u f(u)} \d u \\ &&&= F(u) + c \\ &&&= F(x^m) + c \end{align*}

1. \begin{align*} && \int \frac{1}{u(u-1)} \d u &= \int \left ( \frac{1}{u-1}-\frac{1}{u} \right ) \d u \\ &&&= \ln \left ( \frac{u-1}{u} \right) + c \\ &&&= \ln \left ( 1 - \frac{1}{u} \right) + c \\ && \int \frac{1}{x^n - x} \d x &= \int \frac{1}{x (x^{n-1}-1)} \d x \\ f(u) = u - 1: && &= \frac{1}{n-1} \ln \left ( 1 - \frac{1}{x^{n-1}} \right) + c \end{align*}
2. \begin{align*} v = \sqrt{u+1}, \d v = \tfrac12 (u+1)^{-1/2} \d u && \int \frac{1}{u\sqrt{u+1}} \d u &= \int \frac{1}{(v^2-1)} (u+1)^{-1/2} \d u \\ &&&= \int \frac{2}{v^2-1} \d v \\ &&&=\ln \frac{1-v}{1+v} + c \\ &&&= \ln \left (\frac{1-\sqrt{u+1}}{1+\sqrt{u+1}} \right)+ c \\ f(u) = \sqrt{x+1}:&& \int \frac{1}{\sqrt{x^n + x^2}} \d x &= \int \frac{1}{x\sqrt{x^{n-2}+1}} \d x \\ &&&= \frac{1}{n-2} \ln \left ( \frac{1-\sqrt{x^{n-2}+1}}{1+\sqrt{x^{n-2}+1}} \right)+c \end{align*}

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Solution: \begin{align*} \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right ) &= \frac{1}{\sqrt{1-\left ( \frac{x + a }{ \ x + b} \right )^2}} \left ( \frac{b-a}{(x+b)^2} \right) \\ &= \frac{b-a}{(x+b)\sqrt{(x+b)^2-(x+a)^2}} \\ &= \frac{b-a}{(x+b)\sqrt{(b-a)(2x+b+a)}} \\ &= \frac{\sqrt{b-a}}{(x+b)\sqrt{a+b+2x}} \\ \\ \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right) &= \frac{1}{\sqrt{\left ( \frac{x + b }{ \ x + a} \right)^2-1}} \left ( -\frac{b-a}{(x+a)^2} \right) \\ &= -\frac{b-a}{(x+a)\sqrt{(x+b)^2-(x+a)^2}} \\ &= -\frac{b-a}{(x+a)\sqrt{(b-a)(a+b+2x)}} \\ &= -\frac{\sqrt{b-a}}{(x+a)\sqrt{a+b+2x}} \end{align*}

1. \begin{align*} \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x &= \int \frac{1}{(x+1)\sqrt{\frac12 (2x+6)}} \d x\\ &= \int \frac{\sqrt{2}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= \frac{\sqrt{2}}{2}\int \frac{\sqrt{5-1}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= - \frac{\sqrt{2}}{2}\textrm{arcosh} \left ( \frac{x+5}{x+1} \right) + C \end{align*}
2. \begin{align*} \int \frac{1}{(x+3)\sqrt{x+1}} \d x &= \int \frac{1}{(x+3)\sqrt{\tfrac12(2x+2)}} \d x + C \\ &= \int \frac{\sqrt{3-1}}{(x+3)\sqrt{2x+3-1}} \d x \\ &= \textrm{arcsin} \left ( \frac{x-1}{x+3} \right) \end{align*}

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Solution: \begin{align*} && \arctan a &\in (0, \tfrac{\pi}{4}) \\ && \arctan b &\in (0, \tfrac{\pi}{4}) \\ \Rightarrow && \arctan a+\arctan b &\in (0, \tfrac{\pi}{2}) \\ && \tan \left ( \arctan a+\arctan b \right) &= \frac{\tan \arctan a + \tan \arctan b}{1 - \tan \arctan a \tan \arctan b} \\ &&&= \frac{a+b}{1-ab} \in (0, \infty) \\ \Rightarrow && \arctan \left ( \frac{a+b}{1-ab} \right) &\in (0, \tfrac{\pi}{2}) \\ \Rightarrow && \arctan a + \arctan b &= \arctan \left ( \frac{a+b}{1-ab} \right) \end{align*} Claim: \(\displaystyle \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r\) Proof: (By Induction): Base case (\(n=1\)): \begin{align*} && LHS &= \sum_{r=1}^1 \arctan \left ( \frac{1}{r^2+r+1} \right) \\ &&&= \arctan \left ( \frac{1}{3} \right) \\ && RHS &= \arctan \left ( \frac{1}{1+2} \right)\\ &&&= \arctan \left ( \frac{1}{3} \right) = LHS \end{align*} Inductive step, suppose true for \(n = k\), ie \begin{align*} && \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r &= \arctan \l {k \over k+2} \r \\ \Rightarrow && \sum_{r = 1}^{k+1} \arctan \l {1 \over r^2 + r + 1} \r &= \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r+ \arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \l {k \over k+2} \r+\arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \left ( \frac{{k \over k+2}+\frac{1}{(k+1)^2+(k+1)+1} }{1-\frac{k}{k+2}\frac{1}{(k+1)^2+(k+1)+1} } \right) \\ &&&= \arctan \left ( \frac{k((k+1)^2+k+1+k)+(k+2) }{(k+2)((k+1)^2+(k+1)+1)-k} \right) \\ &&&= \arctan \left ( \frac{k^3+3k^2+4k+2 }{k^3+5k^2+8k+6} \right) \\ &&&= \arctan \left ( \frac{(k+1)(k^2+2k+2) }{(k+3)(k^2+2k+2)} \right) \\ &&&= \arctan \left ( \frac{k+1 }{(k+1)+2} \right) \\ \end{align*} Therefore it is true for \(n = k+1\), therefore it is true for all \(n \geq 1\) by the principle of mathematical induction. \begin{align*} && S &= \lim_{n \to \infty} \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{n}{n+2} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{1}{1+2/n} \r \\ &&&=\arctan\l \lim_{n \to \infty} \frac{1}{1+2/n} \r \\ &&&= \frac{\pi}{4} \end{align*} \begin{align*} && \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r &= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ (r+1)^2 - (r+1) + 1} \right) \\ &&&= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ r^2+r+1} \right) \\ &&&= \arctan \l \frac{1}{0^2+0+1} \r + \frac{\pi}{4} \\ &&&= \frac{\pi}{2} \end{align*}