Problems

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: \begin{align*} && y &= (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \\ && y' &= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + (ax^2+bx+c) \frac{1}{x + \sqrt{1+x^2}} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}} \right) + d\sqrt{1+x^2} + \frac{x(dx+e)}{\sqrt{1+x^2}} \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (ax^2+bx+c) + d(1+x^2) + x(dx+e) \right) \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (a+2d)x^2+(b+e)x+(d+c) \right) \\ \end{align*}

1. We want \(a = 0, b = 1, d = 0, e = -1, c =0\), so \begin{align*} I &= \int \ln \big( x+\sqrt{1+x^2}\,\big) \,\d x \\ &= x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C \end{align*}
2. We want \(a = b =0, e = 0, d = \frac12, c = \frac12\), so \begin{align*} I &= \int \sqrt{1+x^2}\, \d x \\ &= \frac12\ln(x+\sqrt{1+x^2}) + \frac12x\sqrt{1+x^2}+C \end{align*}
3. We want \(a = \frac12, b = 0, d = -\frac14, e = 0, c = \frac14\), so \begin{align*} I &= \int x \ln (x+\sqrt{1+x^2}) \, \d x \\ &= \left (\frac12 x^2+\frac14 \right)\ln(x+\sqrt{1+x^2}) -\frac14x\sqrt{1+x^2}+C \end{align*}

Solution:

1. Let \(\displaystyle y = \frac z {(1+z^2)^{\frac12}}\) then \(\frac{d y}{d x} = \frac{(1+z^2)^{\frac12} - z^2(1+z^2)^{-\frac12}}{1+z^2} = \frac{(1+z^2)-z^2}{(1+z^2)^\frac32} = \frac{1}{(1+z^2)^\frac32}\)
2. \(\kappa = \frac {f''(x)}{\big({1+ (f'(x))^2\big)^{\frac32}}}\) then \begin{align*} && \int \kappa \, dx &= \int \frac{f''(x)}{( 1 + (f'(x))^2)^{\frac32}} \, dx \\ && \kappa x &= \frac{f'(x)}{(1 + (f'(x))^2)^\frac12} + C \\ \Rightarrow && (\kappa x-C)^2 &= \frac{f'(x)^2}{1 + (f'(x))^2} \\ \Rightarrow && f'(x)^2((\kappa x - C)^2 - 1) &= -(\kappa x-C)^2 \\ \Rightarrow && f'(x) &= \frac{\kappa x - C}{\sqrt{1-(\kappa x - C)^2 }} \\ \Rightarrow && f(x) &= \frac{1}{\kappa} \sqrt{1 - (\kappa x - C)^2} \\ \Rightarrow && (\kappa y)^2 + (\kappa x - C)^2 &= 1 \\ \Rightarrow && y^2 + (x - C')^2 &= \frac{1}{\kappa^2} \end{align*} Therefore all the curves are circles and \(\kappa\) is the reciprocal of the radius.

Solution:

1. When \(k =1\), we have \((1-x^2)(y')^2 + y^2 = 1\). Notice that if \(y_1 = x\) we have \(y_1' = 1\) and \((1-x^2) \cdot 1 + x^2 = 1\) so \(y_1\) is a solution, and we are allowed to assume this is the only solution. And notice that \(y_1(1) = 1\).
2. When \(k = 2\) we have \((1-x^2)(y')^2 + 4y^2 = 4\). Trying \(y_2 = 2x^2-1\) we see that \(y_2' = 4x\) and \((1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4\). We can also check that \(y(1) = 2 \cdot 1^2 - 1 = 1\)
3. Let \(z(x) = 2(y_n(x))^2-1\), then \begin{align*} && \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\ \Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2 \\ &&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\ &&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\ &&&= 4n^2 = RHS \end{align*} Therefore \(y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))\) (notice also that \(z(1) = 2(y_n(1))^2-1 = 2-1 = 1\)).
4. Let \(v(x) = y_n(y_m(x))\) so \begin{align*} && (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\ && (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\ \\ && \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\ && (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\ &&&= (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right) \\ &&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\ &&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\ &&&= n^2m^2(1-v^2) \end{align*} Therefore \(v\) satisfies our differential equation and \(v(1) = y_n(y_m(1)) = y_n(1) = 1\) so it must be our desired solution.

Solution: The position of the hands are \(\begin{pmatrix} a\sin(-t) \\ a \cos(-t) \end{pmatrix}\) and \(\begin{pmatrix} b\sin(-60t) \\ b \cos(-60t) \end{pmatrix}\), the distance between the hands is \begin{align*} x &= \sqrt{\left ( a \sin t - b \sin 60t\right)^2+\left ( a \cos t - b \cos 60t\right)^2} \\ &= \sqrt{a^2+b^2-2ab\left (\sin t \sin 60t+\cos t \cos 60t \right)} \\ &= \sqrt{a^2+b^2-2ab \cos(59t)} = \sqrt{a^2+b^2-2ab \cos \theta} \\ \\ \frac{\d x}{\d \theta} &= \frac{ab \sin \theta}{ \sqrt{a^2+b^2-2ab \cos \theta}} \\ \frac{\d^2 x}{\d \theta^2} &= \frac{ab \cos \theta\sqrt{a^2+b^2-2ab \cos \theta} - \frac{a^2b^2 \sin^2 \theta}{\sqrt{a^2+b^2-2ab \cos \theta}} }{a^2+b^2-2ab \cos \theta} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2 \sin^2 \theta }{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2(1-\cos^2 \theta)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab(a^2+b^2) \cos \theta-a^2b^2 \cos \theta- a^2b^2}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{-ab(a\cos \theta -b)(b \cos \theta - a)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ \end{align*} So the rate of increase is largest when \(\cos \theta = \frac{a}{b}\) (since \(\frac{b}{a}\) is impossible. Therefore when \(x = \sqrt{a^2+b^2-2ab \frac{a}{b}} = \sqrt{a^2+b^2-2a^2} = \sqrt{b^2-a^2}\) If \(b = 2a\) then \(\cos \theta = \frac{a}{2a} = \frac12 = \frac{\pi}{3} = 60^\circ\) The relative speed of the hands is \(5.5^\circ\) per minute, so \(\frac{60}{5.5} = \frac{120}{11} \approx 11\) but clearly also less than since \(121 = 11^2\).

Solution:

1. \(\,\) \begin{align*} && f(x) &= f(1-x) \\ \Rightarrow && f'(x) &= -f'(1-x) \\ \Rightarrow && f'(\tfrac12) &= -f'(\tfrac12) \\ \Rightarrow && f'(\tfrac12) &= 0 \\ \\ && f(x) &= f(\tfrac1x) \\ \Rightarrow && f'(x) &= f'(\tfrac1x) \cdot \frac{-1}{x^2} \\ \Rightarrow && f'(-1) &= -f'(-1) \\ \Rightarrow && f'(-1) &= 0 \\ \\ && f'(2) &= -\frac{1}{4}f'(\tfrac12) \\ &&&= 0 \end{align*}
2. Suppose \begin{align*} && f(x) &= \frac{(x^2-x+1)^3}{(x^2-x)^2} \\ && f(1/x) &= \frac{(x^{-2}-x^{-1}+1)^3}{(x^{-2}-x^{-1})^2} \\ &&&= \frac{(1-x+x^2)^3/x^6}{((x-x^2)^2/x^6} \\ &&&= f(x) \\ \\ && f(1-x) &= \frac{((1-x)^2-(1-x)+1)^3}{((1-x)^2-(1-x))^2} \\ &&&= \frac{(1-x+x^2)^3}{(x^2-x)^2} = f(x) \end{align*}
   

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3. Clearly \(x = -1\) is a root of \(f(x) = \frac{27}{4}\), so we must also have \(x=2\) and \(x = \frac12\), therefore \(f(x) > \frac{27}{4}\) if \(x \in \mathbb{R} \setminus \{-1, 2, \tfrac12, 0, 1 \}\). Clearly \(x = 3\) and \(x = -2\) are solutions so we also have: \(\frac13, -\frac12, \frac32, \frac23\) and these must be all solutions so we must have: \(f(x) > \frac{343}{36} \Leftrightarrow x \in (-\infty, -2) \cup (-\frac12, 0) \cup (0, \frac13) \cup (\frac23, 1) \cup (1, \frac32) \cup (3, \infty)\)

Solution: \begin{align*} &&z &= y^n \left( \frac{\d y}{\d x}\right)^{2} \\ \Rightarrow && \frac{\d z}{\d x} &= ny^{n-1}\left( \frac{\d y}{\d x}\right)^{3} + y^{n} \cdot 2 \left( \frac{\d y}{\d x}\right) \left( \frac{\d^2 y}{\d x^2}\right) \\ &&&= y^{n-1} \left( \frac{\d y}{\d x}\right) \left (n \left( \frac{\d y}{\d x}\right)^2 + 2y \frac{\d^2 y}{\d x^2} \right) \end{align*}

1. Let \(z = y (y')^2\), then \begin{align*} && \frac{\d z}{\d x} &= y' \sqrt{y} \\ &&&= \sqrt{z} \\ \Rightarrow && \int z^{-1/2} \d z &= x+C \\ \Rightarrow && 2\sqrt{z} &= x + C \\ x = 0, z=0: && C &= 0 \\ \Rightarrow && y(y')^2 &= \frac14 x^2 \\ \Rightarrow &&\sqrt{y} \frac{\d y}{\d x} &= \frac{1}{2}x\\ \Rightarrow && \int \sqrt{y} \d y &= \int \frac{1}{2}x\d x \\ \Rightarrow && \frac23y^{3/2} &=\frac14x^2 + K \\ x = 0, y = 1: && K &= \frac23 \\ \Rightarrow && y &= \left (\frac38 x^2 + 1 \right)^{2/3} \end{align*}
2. Let \(z = y^{-2} (y')^2\) \begin{align*} && \frac{\d z}{\d x} &= y^{-3} \frac{\d y}{\d x} \left (-2 \left( \frac{\d y}{\d x}\right) + 2y \frac{\d^2 y}{\d x^2} \right) \\ &&&= y^{-3} \frac{\d y}{\d x} 2y^2 \\ &&&= 2y^{-1}(y') = 2 \sqrt{z} \\ \Rightarrow && 2\sqrt{z} &= 2x + C \\ x = 0, z = 0: && C&= 0 \\ \Rightarrow && z &= x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= xy \\ \Rightarrow && \ln |y| &= \frac12 x^2 + K \\ x =0 , y =1; && K &= 0 \\ \Rightarrow && y &= e^{\frac12 x^2} \end{align*}

Solution:

1. \(\,\) \begin{align*} && y &= \ln (x + \sqrt{3+x^2}) \\ \Rightarrow && y' &= \frac{1}{x + \sqrt{3+x^2}} \cdot \left (1 + \frac{x}{\sqrt{3+x^2}} \right) \\ &&&= \frac{1}{\sqrt{3+x^2}} \\ \\ && y &= x\sqrt{3+x^2} \\ && y' &= \sqrt{3+x^2} + \frac{x^2}{\sqrt{3+x^2}} \\ &&&= 2\sqrt{3+x^2} - \frac{3}{\sqrt{3+x^2}} \\ \\ \Rightarrow && \sqrt{3+x^2} &= \frac12(x \sqrt{3+x^2})' + \frac32(\ln(x+\sqrt{3+x^2})' \\ \Rightarrow && \int \sqrt{3+x^2}\, \d x &= \frac12x\sqrt{3+x^2} + \frac32 \ln (x+\sqrt{3+x^2}) + C \end{align*}
2. \(\,\) \begin{align*} && 3 \left ( \frac{\d y}{\d x} \right)^2 + 2x \frac{\d y}{\d x} &= 1 \\ && \frac{\d y}{\d x} &= \frac{-x \pm \sqrt{x^2+3} }3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac16x\sqrt{3+x^2} \pm \frac12 \ln (x+\sqrt{3+x^2}) + C \\ y = 0, x = 1: && 0 &= -\frac16 \pm \frac13 \pm \frac12 \ln 3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac12x\sqrt{3+x^2} \pm \frac32 \ln (x+\sqrt{3+x^2}) + \frac16 \mp \frac13 \mp \frac12 \ln 3 \end{align*}

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*}

Solution: \begin{align*} && y &= \sin(k \sin^{-1} x ) \\ &&y' &= \cos (k \sin^{-1} x) \cdot k \frac{1}{\sqrt{1-x^2}} \\ && y'' &= -\sin (k \sin^{-1} x) \cdot k^2 \frac{1}{(1-x^2)} - \cos(k \sin^{-1} x) \cdot k \frac{x}{(1-x^2)\sqrt{1-x^2}} \\ && (1-x^2)y'' &= -k^2y -xy' \\ \Rightarrow && 0 &= (1-x^2)y''+xy' + k^2y \end{align*} \begin{align*} && y &= Ax^3 + Bx^2 + Cx + D \\ && y' &= 3Ax^2 + 2Bx + C \\ && y'' &= 6Ax+2B \\ && 0 &= (1-x^2)(6Ax+2B) - x( 3Ax^2 + 2Bx + C) + 9(Ax^3 + Bx^2 + Cx + D ) \\ &&&= x^3(-6A-3A+9A) + x^2(-2B-2B+9B) + x(6A-C+9C) + (2B +9D) \\ \Rightarrow && B &= 0 \\ \Rightarrow && D &= 0 \\ \Rightarrow && C &= -\frac34 A \\ \\ x = 0, y = 0, y' = 0: && y &= 3x-4x^3 \\ \end{align*} And so \(\sin 3 x = 3 \sin x - 4\sin^3 x\)

Solution:

1. \begin{align*} && f(x)&=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right) \\ \Rightarrow && f'(x) &= \frac{1}{1+x^2} + \frac{1}{1+\l \frac{1-x}{1+x} \r^2} \cdot \l \frac{-2}{(1+x)^2}\r \\ &&&= \frac1{1+x^2}- \frac{2}{(1+x)^2+(1-x)^2} \\ &&&= \frac1{1+x^2} - \frac{2}{2+2x^2} \\ &&&= 0 \end{align*} Therefore $f(x) = \begin{cases} c_1 & \text{if } x < -1 \\ c_2 & \text{if } x > -1 \end{cases}$ \(f(0) = \tan^{-1} 0 + \tan^{-1} 1 = \frac{\pi}{4}\) \(\lim_{x \to \infty} f(x) = -\frac{\pi}{2} + \tan^{-1} -1 = -\frac{3\pi}{4}\) therefore $f(x) = \begin{cases} -\frac{3\pi}{4}& \text{if } x < -1 \\ \frac{\pi}{4} & \text{if } x > -1 \end{cases}$
2. \begin{align*} && x &= y \sin y^2 \\ \Rightarrow && \frac{\d x}{\d y} &= \sin y^2 + 2y^2 \cos y^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{\sin y^2+2y^2 \cos y^2} \\ &&&=\frac{1}{\frac{x}{y}+2y^2 \sqrt{1-\sin^2y^2}} \\ &&&= \frac{y}{x + 2y^3 \sqrt{1-\frac{x^2}{y^2}}} \\ &&&= \frac{y}{x+2y^2 \sqrt{y^2-x^2}} \end{align*}

Solution: \begin{align*} && y &= x^a \\ && \frac{\d y}{\d x} &= \begin{cases} ax^{a-1} & a \neq 0 \\ 0 & a = 0 \end{cases} \\ \\ && y &= a^x \\ &&&= e^{(\ln a) \cdot x} \\ && \frac{\d y}{\d x} &= \ln a e^{(\ln a) x} \\ &&&= \ln a \cdot a^ x \\ \\ && y &= x^x \\ &&&= e^{x \ln x}\\ && \frac{\d y}{\d x} &= e^{x \ln x} \cdot \left ( \ln x + x \cdot \frac1x \right) \\ &&&= x^x \left (1 + \ln x \right) \\ \\ && y&= x^{(x^x)} \\ &&&= e^{x^ x \cdot \ln x} \\ && \frac{\d y}{\d x} &= e^{x^x \cdot \ln x} \left ( x^x \left (1 + \ln x \right) \cdot \ln x + x^x \cdot \frac1x\right) \\ &&&= x^{x^x} \left (x^x (1+ \ln x) \ln x +x^{x-1} \right) \\ &&&= x^{x^x+x-1} \left (1 + x \ln x + x (\ln x)^2 \right) \\ \\ && y &= (x^x)^x \\ &&&= x^{2x} \\ &&&= e^{2x \ln x} \\ && \frac{\d y}{\d x} &= e^{2 x \ln x} \left (2 \ln x + 2 \right) \\ &&&= 2(x^x)^x(1 + \ln x) \end{align*}

Solution:

1. \begin{align*} && g(x) &= \frac{2x^3+3}{x^4+4} \\ \Rightarrow && g'(x) &= \frac{6x^2(x^4+4) - (2x^3+3)(4x^3)}{(x^4+4)^2} \\ &&&= \frac{-2x^6-12x^3+24x^2}{(x^4+4)} \\ &&&= \frac{-2x^2(x^4+6x-12)}{(x^4+4)} \end{align*} So \(g'(x)\) is not \(0\) for \(x = \pm 1\). We can also note that \(g(-1) = \frac1{5} \neq 1\) Even if the other turning point was \(1\), we would also need to check the behaviour as \(x \to \pm \infty\). We can also note that \(g(-1) = \frac{1}{5} < \frac34\) so the conclusion is also not true.
2. There are several errors. \[ \int (1-x)^{-3} \d x = \underbrace{\frac{1}{4}}_{\text{correct constant}}(1-x)^{-4} + \underbrace{C}_{\text{constant of integration}} \] We cannot integrate through the asymptote at \(1\). There is a sense in which we could argue \(\displaystyle \int_{-1}^3 (1-x)^{-3} \d x = 0\), specifically using Cauchy principal value \begin{align*} \mathrm {p.v.} \int_{-1}^3 (1-x)^{-3} &=\lim_{\epsilon \to 0} \left [ \int_{-1}^{1-\epsilon} (1-x)^{-3} \d x+ \int_{1+\epsilon}^{3} (1-x)^{-3} \d x\right] \\ &=\lim_{\epsilon \to 0} \left [ \left[ \frac14 (1-x)^{-4}\right]_{-1}^{1-\epsilon}+ \left[ \frac14 (1-x)^{-4}\right]_{1+\epsilon}^3\right] \\ &=\lim_{\epsilon \to 0} \left [ \frac14 \epsilon^{-4}-\frac14 \frac1{2^4} + \frac14 \frac1{2^4} - \frac14 \epsilon^{-4} \right] \\ &= \lim_{\epsilon \to 0} 0 \\ &= 0 \end{align*} However, in many normal ways of treating this integral it would be undefined.

Solution: Note that \(z = xy/(x+y+1) \Rightarrow y(x-z) = z(x+1)\) \begin{align*} && g(x) + g(y) &= g(z) \\ \Rightarrow && g'(x) &= g'(z) \cdot \frac{y(x+y+1) - xy \cdot 1} {(x+y+1)^2} \\ &&&= g'(z) \frac{y^2+y}{(x+y+1)^2} \\ &&&= g'(z) \frac{z^2(y^2+y)}{x^2y^2} \\ &&&= g'(z) \frac{z^2(y+1)}{x^2y} \\ &&&= g'(z) \frac{z^2}{x^2} \left (1 + \frac{x-z}{z(x+1)} \right) \\ &&&= g'(z) \frac{z}{x^2} \frac{zx+x}{x+1} \\ &&&= g'(z) \frac{z(z+1)}{x(x+1)} \end{align*} If \(x = 1\) then as \(y\) ranges from \(0\) to \(\infty\), \(z\) ranges from \(0\) to \(1\), so \(g'(1) = \frac{z(z+1)}{1(1+1)}g'(z)\), ie \(2g'(1) = (u^2+u)g'(u)\). \begin{align*} && g'(u) &= \frac{A}{u(u+1)} \\ \Rightarrow && g(u) &= A\int \left ( \frac{1}{u} - \frac{1}{u+1} \right) \d u \\ &&&= A \left ( \ln u - \ln(u+1) \right) + B \\ &&&= A \ln \left ( \frac{u}{u+1} \right) + B \\ \\ && A \ln \left ( \frac{x}{x+1} \right) + B+A \ln \left ( \frac{y}{y+1} \right) + B &=A \ln \left ( \frac{z}{z+1} \right) + B \\ \Rightarrow && B &= A \ln \left ( \frac{z}{z+1} \frac{y+1}{y} \frac{x+1}{x} \right) \\ &&&= A \ln \left ( \frac{1}{1+\frac{x+y+1}{xy}} \frac{(y+1)(x+1)}{xy} \right) \\ &&&= A \ln 1 \\ &&& = 0 \end{align*} Therefore \(B = 0\). \(A\) cannot be determined from \((*)\). Suppose \(f(x) + f(y) = f(z)\), then \(f'(x) = yf'(z)\). Letting \(x = 1\) we find \(f'(1) = uf'(u) \Rightarrow f(u) = C \ln u + D\), but \(D = 0\) so \(f(x) = C \ln x\)