Problems

---

Solution:

1. \begin{align*} && S + T &= \int \frac{\cos x + \sin x }{\cos x + \sin x} \d x \\ &&&= \int \d x \\ &&&= x + C \\ && S - T &= \int \frac{\cos x - \sin x}{\cos x + \sin x} \d x \\ &&&= \ln( \cos x + \sin x) + C \\ \Rightarrow && 2S &= x + \ln(\cos x + \sin x) + C \\ \Rightarrow && S &= \frac12 \left ( x + \ln(\cos x + \sin x) \right) + C \\ \Rightarrow && 2T &= x - \ln(\cos x + \sin x) + C \\ \Rightarrow && T &= \frac12 \left ( x - \ln(\cos x + \sin x) \right) + C \end{align*}
2. \begin{align*} && I &= \int_{1/4}^{1/2} (1-4x)\sqrt{\frac1x-1} \d x \\ x = \sin^2 t, \d x = 2 \sin t \cos t \d t: &&&= \int_{\pi/6}^{\pi/4} (1-4\sin^2 t) \sqrt{\frac{1-\sin^2 t}{\sin^2 t}} 2 \sin t \cos t \d t\\ &&&=\int_{\pi/6}^{\pi/4} (1-4\sin^2 t)\frac{\cos t}{\sin t} 2 \sin t \cos t \d t \\ &&&= \int_{\pi/6}^{\pi/4} (1-4\sin^2 t) 2 \cos^2 t \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 2\cos^2t - 8 \sin^2t \cos^2 t \right) \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 1+\cos2t - 2 \sin^2 2t \right) \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 1+\cos2t +(\cos 4t-1)\right) \d t \\ &&&= \left[\frac12 \sin 2t + \frac14 \sin 4t \right]_{\pi/6}^{\pi/4} \\ &&&= \left ( \frac12 \right) - \left (\frac12 \frac{\sqrt{3}}{2} + \frac14 \frac{\sqrt{3}}{2} \right) \\ &&&= \frac{4-3\sqrt{3}}{8} \end{align*}

---

Solution: \begin{align*} f(x) & =nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n} \\ f'(x) &= n - \binom{n}{2} x + \binom{n}{3}x^2 - \cdots (-1)^{r+1} \binom{n}{r} + \cdots + (-1)^{n+1} x^{n-1} \\ &= \frac{1-(1-x)^n}{x} \end{align*} Therefore, since \(\displaystyle f(x) = \int_0^xf'(t)\,dt\) \begin{align*} f(x) &= \int_0^x \frac{1 - (1-t)^n}{t} \, dt \\ &= \int_{1}^{1-x} \frac{1-y^n}{1-y} (-1)\, dy \tag{Let \(y = 1-t, \frac{dy}{dt} = -1\)} \\ &= \boxed{\int_{1-x}^1 \frac{1-y^n}{1-y} dy} \\ &= \int_{1-x}^1 \l 1 + y + y^2 + \cdots + y^{n-1} \r \, dy \\ &= \left [ y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots + \frac{y^n}{n} \right]_{1-x}^1 \\ \end{align*} So when \(x = 1, 1-x = 0\) so we exactly have the sum required.

---

Solution:

1. \(,\) \begin{align*} u = 1/y, \frac{\d u}{\d x} = -1/y^2 y': && -\frac{\d u}{\d x} + u &= e^{2x} \\ \Rightarrow && \frac{\d}{\d x} \left (e^{-x}u \right) = -e^{x} \\ \Rightarrow && e^{-x}u &= -e^x+C \\ \Rightarrow && u &= Ce^x-e^{2x} \\ \Rightarrow && y &= \frac{1}{Ce^x-e^{2x}} \end{align*}
2. \(\,\) \begin{align*} u = 1/y^2, u' = -2/y^3 y': && -2u'+u &= e^{2x} \\ \Rightarrow && \frac{\d}{\d x} \left (e^{-1/2x}u\right) &= -\frac12e^{\frac32x} \\ \Rightarrow && e^{-\frac12x}u &= -\frac13e^{\frac32x}+C \\ \Rightarrow && u &= -\frac13e^{2x}+Ce^{\frac12x} \\ \Rightarrow && y &= \left ( \frac{1}{Ce^{\frac12x}-\frac13e^{2x}}\right)^{\frac12} \end{align*}

---

Solution: \begin{align*} && \dot{x} &= (2K-x)x \\ \Rightarrow && \int \d t &= \int \frac{1}{(2K-x)x} \d x \\ &&&= \int \frac1{2K}\left ( \frac{1}{2K-x} + \frac{1}{x} \right) \d x \\ &&&= \frac{1}{2K} \left (\ln x - \ln (2K-x) \right) \\ \Rightarrow && 2Kt+C &= \ln \frac{x}{2K-x} \\ t = 0, x = K: && C &= \ln \frac{K}{2K-K} = 0 \\ \Rightarrow && e^{2Kt} &= \frac{x}{2K-x} \\ \Rightarrow && e^{-2Kt} &= \frac{2K}{x} -1 \\ \Rightarrow && x &= \frac{2K}{1+e^{-2Kt}} \\ &&&= K + K \frac{1-e^{-2Kt}}{1+e^{-2Kt}} \end{align*}

As \(t \to \infty\) \(x(t) \to 2K\) We now have \[ \dot{x} = (2K-x)x - L\] Suppose \(y = x - K + \alpha\), where \(\alpha = \sqrt{K^2-L}\) then, \begin{align*} && \dot{y} &= \dot{x} \\ &&&= (2K-x)x - L \\ &&&= (2K - (y+K-\alpha))(y+K-\alpha) - L \\ &&&= (K+\alpha - y)y + (K+\alpha-y)(K-\alpha) - L \\ &&&= (K+\alpha-y)y + K^2-\alpha^2 -y(K-\alpha) - L \\ &&&= (K+\alpha-y)y -y(K-\alpha) \\ &&&= (2\alpha-y)y \end{align*} Note that when \(t = 0, x = K, y = \alpha\) so we have the original equation but with \(x \to y\) and \(\alpha \to K\), in particular as \(t \to \infty\) \(y \to 2\alpha\) and \(x \to 2\alpha - \alpha + K = K +\alpha\)

---

Solution: \begin{align*} && I &= \int_0^{\frac14\pi} \ln (1 + \tan \theta) \d \theta \\ \theta = \tfrac14\pi - \phi, \d \theta = -\d\phi: &&&= \int_0^{\frac14 \pi} \ln ( 1 + \tan (\tfrac14\pi - \phi)) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( 1 + \frac{1 - \tan \phi}{1+\tan \phi} \right) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( \frac{2}{1+\tan \phi} \right) \d \phi \\ &&&= \tfrac14 \pi \ln 2 - I \\ \Rightarrow && I &= \tfrac18\pi \ln 2 \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln(1+x)}{1+x^2} \d x \\ x= \tan \theta \d \theta, \d \theta = \frac{\d x}{1+x^2} &&&= \int_0^{\frac14 \pi} \ln(1 + \tan \theta) \d \theta \\ &&&= \tfrac18 \pi \ln 2 \end{align*} \begin{align*} && K &= \int_0^{\frac12 \pi} \ln \left ( \frac{1 + \sin x}{1 + \cos x} \right) \d x \\ y = \tfrac12\pi - x, \d y = -\d x: &&&= \int_0^{\frac12\pi} \ln \left ( \frac{1+\cos y}{1+\sin y}\right) \d y \\ &&&= -K \\ \Rightarrow && K &= 0 \end{align*}

---

Solution:

1. \begin{align*} && x^{4}+10x^{3}+26x^{2}+10x+1 &= 0 \\ \Leftrightarrow && x^2 + 10x + 26 + 10x^{-1} + x^{-2} &= 0 \\ \Leftrightarrow && (x^2 + x^{-2} + 2) + 10(x+x^{-1}) + 24 &= 0 \\ \Leftrightarrow && y^2 + 10y + 24 &= 0 \tag{\(y = x + x^{-1}\)} \\ \Leftrightarrow && (y+6)(y+4) &= 0 \\ \Leftrightarrow && \begin{cases} x+x^{-1} = -4 \\ x+x^{-1} = -6 \\ \end{cases} \\ \Leftrightarrow && \begin{cases} x^2+4x+1 = 0 \\ x^2+6x+1 = 0 \\ \end{cases} \\ \Leftrightarrow && \boxed{\begin{cases} x = -2 \pm \sqrt{3} \\ x = -3 \pm 2\sqrt{2} \\ \end{cases}} \\ \end{align*}
2. \begin{align*} && x^{4}+x^{3}-10x^{2}-4x+16=0 &= 0 \\ \Leftrightarrow && x^2 + x - 10 - 4x^{-1} + 4x^{-2} &= 0 \\ \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1}) - 6 &= 0 \\ \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1}) - 6 &= 0 \\ \Leftrightarrow && z^2 + z - 6 &= 0 \tag{\(z = x -2x^{-1}\)} \\ \Leftrightarrow && (z+3)(z-2) &= 0 \\ \Leftrightarrow && \begin{cases} x-2x^{-1} = -3 \\ x-2x^{-1} = 2 \\ \end{cases} \\ \Leftrightarrow && \begin{cases} x^2+3x-2 = 0 \\ x^2-2x-2 = 0 \\ \end{cases} \\ \Leftrightarrow && \boxed{\begin{cases} x = \frac{-3 \pm \sqrt{17}}{2} \\ x = 1 \pm \sqrt{3} \\ \end{cases}} \\ \end{align*}

---

Solution: \begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ t = \pi - x, \d t = -\d t : &&&= \int_{t = \pi}^{t = 0} (\pi - t) f(\sin (\pi - t)) -\d t \\ &&&= \int_0^{\pi} (\pi - t) f(\sin t) \d t \\ \Rightarrow && 2 I &= \pi \int_0^\pi f(\sin t) \d t \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \d x \end{align*} \begin{align*} && I &= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \frac{\pi}{2}\int_0^\pi \frac{\sin x}{1 + \cos^2 x} \d x \\ &&&= \frac{\pi}{2}\left [ -\tan^{-1} \cos x\right]_0^{\pi} \\ &&&= \tan 1 - \tan (-1) = \frac{\pi^2}{4} \\ \\ && I &= \int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x + \int_{\pi}^{2\pi} \frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ u = x - \pi, \d u = \d x: &&&= \frac{\pi^2}{4} + \int_{0}^{\pi} \frac{(u+\pi)(-\sin u)}{1 + \cos^2 u}\d u \\ &&&= \frac{\pi^2}{4} -\frac{3\pi}{2} \int_0^{\pi} \frac{\sin u}{1+\cos^2 u} \d u \\ &&&= - \frac{\pi^2}2 \end{align*}

---

Solution: \begin{align*} && I &= \int_0^1 \cos^{-1} t \d t \\ \cos y = t: -\sin y \d y = \d t: &&&= \int_{\frac{\pi}{2}}^0 -y \sin y \d y \\ &&&= \int_0^{\pi/2} y \sin y \d y \\ &&&= \left [-y \cos y \right]_0^{\pi/2} + \int_0^{\pi/2} \cos y \d y \\ &&&= \left [ \sin y \right]_0^{\pi/2} = 1 \end{align*}

If \(X = x\) then the rod will cross the line if \(\frac{x}{\sin \theta} < a\) or \(\frac{2b-x}{\sin \theta} < a\), ie \(a\sin \theta > \max (x, 2b-x)\). Therefore the probability is \(\frac{2\sin^{-1} \left (\max(\frac{x}{a}, \frac{2b-x}{a}) \right)}{\pi}\). Therefore the probability the pin crosses a line is: \begin{align*} \mathbb{P} &= \frac{1}{2b}\int_{x=0}^{x=2b} \frac{2\sin^{-1} \left (\max(\frac{x}{a}, \frac{2b-x}{a}) \right)}{\pi} \d x \\ &= \frac{2a}{b\pi} \end{align*}

---

Solution:

1. \begin{align*} y = \pi - x, \d y = -\d x: && \int_0^{\pi} x f(\sin x) &= \int_{y = \pi}^{y = 0}(\pi - y) f(\sin(\pi-y))- \d y \\ &&&= \int_0^{\pi} (\pi -y) f(\sin y) \d y \\ \Rightarrow && 2 \int_0^{\pi} x f(\sin x)\d x &= \pi \int_0^{\pi} f(\sin x) \d x \\ &&&= \pi \int_0^{\pi/2} f(\sin x ) \d x + \pi \int_{\pi/2}^{\pi} f(\sin x ) \d x \\ &&&= \pi \int_0^{\pi/2} f(\sin x ) \d x +\pi \int_{y=\pi/2}^{y=0} f(\sin (\pi-y) ) (-\d y) \\ &&&= 2 \pi \int_0^{\pi/2} f(\sin x) \d x \\ \Rightarrow && \int_0^{\pi} x f(\sin x)\d x &= \pi \int_0^{\pi/2} f(\sin x) \d x \end{align*} Therefore if \(f(x) = \frac1{2+\sin x}\), letting \(t = \tan \frac{x}{2}\) we have \(\sin x = \frac{2 t}{1+t^2}, \frac{dt}{\d x} = \frac12 (1+t^2)\) \begin{align*} && \int_0^{\pi} \frac{x}{2 + \sin x } \d x &= \pi \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\ &&&= \pi \int_{t = 0}^{t = 1} \frac{1}{2+\frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\ &&&=\pi \int_0^1 \frac{2}{2t^2+2t+2} \d t\\ &&&=\pi \int_0^1 \frac{1}{(t+\tfrac12)^2 + \tfrac34} \d t\\ &&&= \pi \left [\frac{1}{\sqrt{3/4}} \tan^{-1} \frac{u}{\sqrt{3/4}} \right ]_{u=1/2}^{3/2} \\ &&&= \frac{2 \pi}{\sqrt{3}} \left ( \tan^{-1} \sqrt{3} - \tan^{-1} \frac1{\sqrt{3}} \right) \\ &&&= \frac{2 \pi}{\sqrt{3}} \left ( \frac{\pi}{3} - \frac{\pi}{6} \right) \\ &&&= \frac{\pi^2}{3\sqrt{3}} \end{align*}
2. Let \(u = (\sin^{-1} t)^2, \frac{\d u}{\d t} = 2(\sin^{-1} t) \frac{1}{\sqrt{1-t^2}}\) \begin{align*} \int_{0}^{1}\frac{(\sin^{-1}t)\cos\left[(\sin^{-1}t)^{2}\right]}{\sqrt{1-t^{2}}}\,\mathrm{d}t &= \int_{u=0}^{\pi^2/4} \frac12 \cos u \d u \\ &= \frac12 \sin \frac{\pi^2}{4} \end{align*}

---

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 \(f(x) = C \ln x\)

---

Solution: \begin{align*} && 0 &= u^2 + 2u \sinh x -1 \\ &&&= u^2 + u(e^x-e^{-x})-e^{x}e^{-x} \\ &&&= (u-e^{-x})(u+e^x) \\ \Rightarrow && u &= e^{-x}, -e^x \end{align*} \begin{align*} && 0 &= \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1 \\ \Rightarrow && \frac{\d y}{\d x} &= e^{-x}, -e^x \\ \Rightarrow && y &= -e^{-x}+C, -e^x+C \\ y(0) = 0: && C &= 1\text{ both cases } \\ y'(0) > 0: && y &= 1-e^{-x} \end{align*} \begin{align*} && 0 &= \sinh x u^2 + 2u -\sinh x \\ \Rightarrow && u &= \frac{-2 \pm \sqrt{4+4\sinh^2 x}}{2\sinh x} \\ &&&= \frac{-1 \pm \cosh x}{\sinh x} = - \textrm{cosech }x \pm \textrm{coth}x \\ \\ && 0 &= \sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x \\ \Rightarrow && \frac{\d y}{\d x} &= - \textrm{cosech }x \pm \textrm{coth}x \\ \Rightarrow && y &= -\ln \left ( \tanh \frac{x}{2} \right) \pm \ln \sinh x+C \end{align*} For \(x \to 0\) to be defined, we need \(+\), so \begin{align*} && y &= \ln \left (\frac{\sinh x}{\tanh \frac{x}{2}} \right) + C \\ && y &= \ln \left (\frac{2\sinh \frac{x}{2} \cosh \frac{x}{2}}{\tanh \frac{x}{2}} \right)+C \\ &&&= \ln \left (2 \cosh^2 x \right) + C \\ y(0) = 0: && 0 &= \ln 2+C \\ \Rightarrow && y &= \ln(2 \cosh^2 x) -\ln 2 \\ && y &= 2 \ln (\cosh x) \end{align*}

---

Solution:

1. \begin{align*} u = \frac{\pi}{2} - x :&& \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_{\frac12\pi}^0 \ln (\cos u) (- 1)\d u \\ &&&= \int_0^{\frac12 \pi} \ln (\cos x) \d x \\ \Rightarrow && 2 \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_0^{\tfrac12 \pi} \ln (\sin x) \d x +\int_0^{\tfrac12 \pi} \ln (\cos x) \d x \\ &&&= \int_0^{\tfrac12 \pi}\left (\ln (\sin x)+ \ln (\cos x) \right) \d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\frac12 \sin 2x \right) \d x \\ &&&= \int_0^{\frac12 \pi} \left ( \ln \left (\sin 2x \right) - \ln 2 \right)\d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{2} \ln 2\\ \Rightarrow && \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \frac12 \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{4} \ln 2 \end{align*} \begin{align*} u = 2x, \d u = 2 \d x && \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x &= \int_0^{\pi} \ln (\sin u) \frac12 \d u \\ &&&= \frac12 \int_0^{\pi} \ln (\sin u) \d u \\ &&&=\frac12 \left ( \int_0^{\pi/2} \ln (\sin u) \d u + \int_{\pi/2}^{\pi} \ln (\sin u) \d u \right)\\ &&&= \int_0^{\pi/2} \ln (\sin u) \d u \\ \Rightarrow && I &= \frac12 I - \frac14 \pi \ln 2 \\ \Rightarrow && I &= -\frac12 \pi \ln 2 \end{align*}
2. \begin{align*} && \ln u &< u & \quad (u \geq 1)\\ \underbrace{\Rightarrow}_{u = \sqrt{x}} && \ln \sqrt{x} &< \sqrt{x} \\ \Rightarrow && \frac12 \ln x &< \sqrt{x} \\ \Rightarrow && \frac{\ln x}{x} &< \frac{2\sqrt{x}}{x} \\ &&&= \frac{2}{\sqrt{x}} \\ &&&\to 0 & (x \to \infty) \\ && x \ln x &= \frac{\ln 1/y}{y} \\ &&&= -\frac{\ln y}{y} \\ &&&\to 0 & (y \to \infty, x \to 0) \end{align*}
3. \begin{align*} \int_{0}^{\frac{1}{2}\pi}x\cot x\,\mathrm{d}x &= \left [ x \ln(\sin x) \right]_0^{\pi/2} - \int_0^{\pi/2} \ln (\sin x) \d x \\ &= \left ( \frac{\pi}{2} \ln 1 - \lim_{x \to 0} x \ln (\sin x) \right) - \left ( -\frac12 \pi \ln 2 \right) \\ &= \frac12 \pi \ln 2 \end{align*}

---

Solution: \[ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \]

1. \begin{align*} (1+1)^n &= \sum_{k=0}^n \binom{n}{k} \\ (1-1)^n &= \sum_{k=0}^n (-1)^n\binom{n}{k} \\ &= \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} -\sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} \end{align*} Therefore \(\displaystyle \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} = \sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} = \frac{2^n}{2} = 2^{n-1}\)
2. \begin{align*} && (1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{k=0}^n k\binom{n}{k} x^{k-1} \\ x = 1: && n2^{n-1} &= \sum_{k=1}^n k\binom{n}{k} \end{align*} as required

\begin{align*} \sum_{r=0}^n (r + (-1)^r) \binom{n}{r} &= n2^{n-1}+0 = n2^{n-1} \end{align*}