Problems

---

Solution:

1. \(\,\) \begin{align*} && \int_{1/m}^m \frac{x^2}{x+1} \d x &= \int_{1/m}^m \left ( x- 1 + \frac{1}{x+1} \right) \d x \\ &&&= \left [ \frac{x^2}{2} - x + \ln (x+1) \right]_{1/m}^m \\ &&&= \left ( m^2/2 - m + \ln(m+1) \right)- \left ( \frac{1}{2m^2} - \frac{1}{m} + \ln\left(\frac1m+1\right) \right) \\ &&&= \frac{m^4-2m^3-1+2m}{2m^2} + \ln (m+1) - \ln(m+1) + \ln m \\ &&&= \frac{(m-1)^3(m+1)}{2m^2} + \ln m \end{align*}
2. \(\,\) \begin{align*} u = \frac{1}x, \d x = -\frac{1}{u^2} \d u:&& \int_{1/m}^m \frac1 {x^n(x+1)}\,\d x &= \int_{u=m}^{u=1/m} \frac{1}{u^{-n}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/m}^m \frac{u^{n-1}}{u+1} \d u \end{align*}
3. • \(\bf (a)\) \(\,\) \begin{align*} && I &= \int_{1/2}^2 \frac {x^5+3}{x^3(x+1)}\,\d x \\ &&&= \int_{1/2}^2 \left ( \frac{x^2}{x+1} + \frac{3}{x^3(x+1)} \right) \d x \\ &&&= \int_{1/2}^2 \frac{x^2}{x+1} \d x + 3 \int_{1/2}^2 \frac{x^2}{x+1} \d x \\ &&&= 4 \left ( \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \right) \\ &&&= \frac32+4 \ln 2 \end{align*}
   • \(\bf (b)\) \(\,\) \begin{align*} && J &= \int_1^2 \frac{x^5+x^3 +1}{x^3(x+1)}\, \d x \\ && K &= \int_1^2 \frac{x^5 +1}{x^3(x+1)}\, \d x\\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=1}^{u=1/2} \frac{u^{-5}+1}{u^{-3}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/2}^1 \frac{1 + u^5}{u^3(u+1)} \d u \\ \Rightarrow && K &= \frac12 \int_{1/2}^2 \frac{x^5+1}{x^3(x+1)} \d x \\ &&&= \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \\ &&&= \frac38 + \ln 2 \\ && L &= \int_1^2 \frac{x^3}{x^3(x+1)} \d x \\ &&&= \ln (3) - \ln 2 \\ \Rightarrow && J &= \frac38 + \ln 3 \end{align*}

---

Solution: \begin{align*} && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\ \Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\ \Rightarrow && 2 &= 2B \tag{\(x = 1\)} \\ \Rightarrow && 1 &= B \\ \Rightarrow && 1 &= A+B+D \tag{\(x = 0\)}\\ \Rightarrow && A &= -D \\ \Rightarrow && 0 &= 4A+2B-4C+4D \tag{\(x = -1\)}\\ \Rightarrow && C &= \frac12\\ \Rightarrow && 3 &= -5A+5B+2C+D \tag{\(x=2\)} \\ \Rightarrow && 3 &= -6A+6 \\ \Rightarrow && A,D &=-\frac12,\frac12 \\ \Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\ &&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k} \end{align*} Therefore the coefficient of \(x^n\) is \(n+1\) or \(n+2\) depending on whether the coefficients from the final series add constructively \(n \equiv 1, 2 \pmod{4}\) or destructively. \begin{align*} \frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\ &= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\ &= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\ & \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\ &= 1.34457890 + \frac{12}{10^{10}} + \cdots \end{align*} \begin{align*} && \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\ && &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\ \Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\ &&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right ) \end{align*} Therefore for this will be less than \(10^{-9}\), when \(m = 11\), so our approximation is valid to 9sf

---

Solution: \begin{align*} && y &= \frac{2x(x^2-5)}{x^2-4} \\ &&&= 2x(x^2-5)(-\tfrac14)(1-\tfrac14x^2)^{-1} \\ &&&= \tfrac52x + \cdots \\ &&&= \frac{2x(x^2-4)-2x}{x^2-4} \\ &&&= 2x - \frac{2x}{x^2-4} \end{align*}

1. We are looking for the intersections of \(y = \frac23(x+3)\) and \(y = f(x)\)
   

   + - ↺
   Therefore 3 real roots.
2. We are looking for intersections of \(y = \frac12(5x-2)\) and \(y = f(x)\)
   

   + - ↺
   so one solution.
3. We are looking for intersections of \(y = f(x)^2\) and \(y = x^2+1\), or \(y = \sqrt{x^2+1}\) and \(y = f(x)\) where \(f(x) \geq 0\)
   

   + - ↺
   So \(3\) solutions.

---

Solution:

1. Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
2. \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
3. 

   + - ↺

---

Solution:

1. \(\,\) \begin{align*} && I &= \int_0^1 \frac 1 {x^2 + (a+2)x +2a} \; \d x \\ &&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x\\ \end{align*} Case 1: \(a = 2\) \begin{align*} && I &= \int_0^1 \frac{1}{(x+2)^2} \d x \\ &&&= \left [ -(x+2)^{-1}\right]_0^1 = \frac12 - \frac13 = \frac16 \end{align*} Case 2: \(a \neq 2, a \not \in [0,1]\) \begin{align*} && I &=\frac{1}{a-2} \int_0^1 \left ( \frac{1}{x+2} - \frac{1}{x+a} \right) \d x \\ &&&= \frac{1}{a-2} \left [ \ln |x+2| - \ln |x + a|\right]_0^1 \\ &&&= \frac{1}{a-2} \left ( \ln \frac{3}{|1+a|} - \ln \frac{2}{|a|} \right) \\ &&&= \frac{1}{a-2} \ln \frac{3|a|}{2|a+1|} \end{align*} Case 3: \(a \in [0, 1]\), \(I\) does not converge
2. \(\,\) \begin{align*} && J &= \int _{1}^2\frac 1 {u^2 +au +a-1} \; \d u \\ &&&= \int_1^2 \frac{1}{(u+a-1)(u+1)} \d u \\ x = u-1:&&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x \end{align*} So it's the same as the previous integral

---

Solution: If \(m\) is even, clearly that expression is positive since it's the sum of two (different) squares. If \(m\) is odd, then we can expand it as a sum of powers of \(x^2\) with a leading coefficient of \(1\) so it is also positive. \begin{align*} && f (x) = \frac{ (1+x )^m - ( 1-x )^m}{ (1+x )^m + (1-x )^m} \\ && f'(x) &= \frac{(m(1+x )^{m-1} + m( 1-x )^{m-1})((1+x)^m + (1-x)^m ) - ((1+x )^m - ( 1-x )^m)(m(1+x)^{m-1} - m(1-x)^{m-1} )}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^m(1-x)^{m-1}+2m(1+x)^{m-1}(1-x)^m}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^{m-1}(1-x)^{m-1}(1+x+1-x)}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{4m(1+x)^{m-1}(1-x)^{m-1}}{\l (1+x)^m + (1-x)^m \r^2} \\ \end{align*}

---

Solution:

\begin{align*} && y &= (1+ x^2)^{-1} \\ \Rightarrow && y' &= -2x(1+x^2)^{-2} \\ \text{eqn of tangent}:&& \frac{y - (1+t^2)^{-1}}{x-t} &= -2t(1+t^2)^{-2} \\ \text{passes thru }(0,1): && \frac{1-(1+t^2)^{-1}}{-t} &= -2t(1+t^2)^{-2} \\ \Rightarrow && (1+t^2)^2-(1+t^2) &= 2t^2 \\ \Rightarrow && t^4-t^2 &= 0 \\ \Rightarrow && t &= 0, \pm 1 \\ \Rightarrow && \frac{y - \frac12}{x - 1} &= -\frac12 \\ && y &=1 -\tfrac12 x \end{align*} There can be no further intersections since the equation is equivalent to the cubic \((1-\frac12 x)(1+x^2) = 1\) and we have already found \(3\) roots. \begin{align*} && A &= \int_0^1 \frac{1}{1 + x^2} = \frac{\pi}{4} \\ && A &> \frac12 \cdot 1 \cdot (1 + \tfrac12) = \frac34 \\ \Rightarrow && \pi &> 3 \\ \\ && V &=\pi \int_{\frac12}^1 x^2 \d y \\ &&&= \pi \int_{\frac12}^1 \left ( \frac{1}{y}-1 \right) \d y \\ &&&= \pi \left [\ln y \right]_{1/2}^1-\frac12 \\ &&&= \pi \ln 2 - \frac{\pi}{2} \\ && V &> \frac13 \pi 1^2 \frac{1}{2} \\ &&&= \frac{\pi}{6} \\ \Rightarrow && \ln 2 &> \frac{2}{3} \end{align*}

---

Solution:

1. For the first one, consider \begin{align*} && \int_0^1 \frac{1}{(1+tx)^2} \d x &= \left [ -\frac{1}{t}(1+tx)^{-1} \right]_0^1 \\ &&&= \frac{1}{t} - \frac{1}{t(1+t)} \\ &&&= \frac{t+1-1}{t(t+1)} = \frac{1}{t+1} \end{align*}
2. Consider \begin{align*} && \int_0^1 \frac{-2x}{(1+tx)^3} \d x &= \int_0^1 \frac{\frac{2}{t}(1+tx) -\frac{2}{t}}{(1+tx)^3} \d x \\ &&&= -\frac{2}{t} \int_0^1 \left (\frac{1}{(1+tx)^2}- \frac{1}{(1+tx)^3} \right) \d x \\ &&&= -\frac{2}{t} \frac{1}{t+1} + \frac{2}{t} \left [ \frac{1}{-2t}(1+tx)^{-2}\right]_0^1 \\ &&&= -\frac{2}{t(t+1)} + \frac2t\left (\frac{1}{2t} - \frac{1}{2t(1+t)^2} \right) \\ &&&= -\frac{2}{t} \left ( \frac{1}{t+1} + \frac{1}{2t(1+t)^2} - \frac{1}{2t}\right) \\ &&&= -\frac{2}{t} \frac{2t(1+t)+1-(1+t)^2}{2t(1+t)^2} \\ &&&= -\frac{2}{t} \frac{2t^2+2t+1-1-2t-t^2}{2t(1+t)^2} \\ &&&= -\frac{1}{(1+t)^2} \end{align*}

I would expect it to be \(\frac{2}{(1+t)^3}\). This is actually an application of differentiating under the integral sign and is completely valid where functions are well behaved.

---

Solution:

1. \(x^2 + 1 \geq 2x \Rightarrow \frac{1}{x^2+1} \leq \frac1{2x} \Rightarrow \frac{x^6}{(x^2+1)^4}\leq \frac{x^6}{16x^2} = \frac1{16}x^2 \leq \frac1{16}\) with equality when \(x = 1\)
2. \(\,\) \begin{align*} && RHS &= \frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \\ &&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)^3-6x(x^2+1)^2(Ax^5+Bx^3+Cx)}{(x^2+1)^6}+\frac{Dx^6}{(x^2+1)^4} \\ &&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)-6x(Ax^5+Bx^3+Cx)+Dx^6}{(x^2+1)^4} \\ &&&= \frac{(D-A) x^6 + (5A-3B) x^4 + (3B-5C)x^2 + C}{(x^2+1)^4} \\ \Rightarrow && C &= 1 \\ && 3B &= 5 \quad\quad\quad\Rightarrow B = \frac53 \\ && 5A &= 3B = 5\Rightarrow A = 1 \\ && D &= A \quad\quad \Rightarrow D = 1 \end{align*}
3. So \begin{align*} && I &= \int_{0}^{1}\frac{1}{(x^{2}+1)^{4}}\, \d x \\ &&&= \int_{0}^{1}\frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \, \d x \\ &&&= \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1 + \int_{0}^{1}\frac{Dx^6}{(x^2+1)^4} \, \d x \\ &&& \leq \frac{A+B+C}{8} + \frac1{16} \\ &&&= \frac{2+\frac53}{8} + \frac1{16} = \frac{11}{24} + \frac{1}{16} \\ && I &\geq \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1 = \frac{11}{24} \end{align*}

---

Solution: \begin{align*} && f(x) &= a x-\frac{x^{3}}{1+x^{2}} \\ \Rightarrow && f'(x) &= a - \frac{3x^2(1+x^2)-x^3 \cdot 2 x}{(1+x^2)^2} \\ &&&= a - \frac{-x^4+3x^2}{(1+x^2)^2} \\ &&&= a - \frac{-t^2+3t}{(1+t)^2} \\ &&&= \frac{a+2at+at^2-t^2-3t}{(1+t)^2} \\ &&&= \frac{(a-1)t^2+(2a-3)t+a}{(1+t)^2} \\ \\ && 0 \leq \Delta &= (2a-3)^2 - 4 \cdot (a-1) \cdot a \\ &&&= 4a^2-12a+9 - 4a^2+4a \\ &&&= -8a + 9 \\ \Leftrightarrow && a &\geq 9/8 \end{align*} Therefore if \(a \geq 9/8\) the numerator is always non-negative and \(f'(x) \geq 0\)

---

Solution: \begin{align*} && x+ c &= f(x) \\ \Rightarrow && (x+c)(x^2-1) &= x(x-2)(x-a) \\ \Rightarrow && x^3 + cx^2-x-c &= x^3-(2+a)x^2+2ax \\ \Rightarrow && 0 &= (c+2+a)x^2-(1+2a)x-c \\ && 0 &\leq \Delta = (1+2a)^2 + 4(2+c+a)c \\ &&&= 4c^2+(4a+8)c + (1+2a)^2 \\ && \Delta_c &= 16(a+2)^2-16(1+2a)^2 \\ &&&= 16(1-a)(3a+3) \\ &&&= 48(1-a^2) \end{align*} Therefore if \(|a| \geq 1\) we must have \(\Delta_c \leq 0\) which means \(\Delta \geq 0\) and so there are always solutions. If \(|a| < 1\) there are values for \(c\) where \(\Delta < 0\) and there would be no solutions. \begin{align*} && y &= \frac{x^3-(2+a)x^2+2ax}{x^2-1} \\ &&&= \frac{(x^2-1)(x-(2+a))+(2a+1)x-(2+a)}{x^2-1} \\ &&&= x - (2+a) + \frac{(2a+1)x-(2+a)}{x^2-1} \end{align*} therefore the oblique asymptote has equation \(y = x - (2+a)\)

1. 

   + - ↺
2. 

   + - ↺

---

Solution: \begin{align*} && y &= \frac{ax+b}{cx+d} \\ &&&= \frac{\frac{a}{c}(cx+d) - \frac{da}{c} + b}{cx+d} \\ \Rightarrow && y' &= \left (b - \frac{da}{c} \right)(-1)(cx+d)^{-2} \cdot c \\ &&&= (ad-bc)(cx+d)^{-2} \end{align*} \begin{align*} && y &= \frac{x+1}{x+3} \\ && \frac{\d y}{\d x} &= \frac{2}{(x+3)^2} \\ \Rightarrow && I &= \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x+1}{x+3}\right)\d x \\ &&&= \int_{y=1/3}^{y=1/2} \frac12 \ln y \, \d y \\ &&&= \frac12 \left [ y \ln y - y \right]_{1/3}^{1/2} \\ &&&= \frac12 \left ( \frac12\ln \frac12 - \frac12 - \frac13 \ln\frac13 + \frac13 \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} \end{align*} \begin{align*} && J &= \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x^2+3x+2}{(x+3)^2} \right) \d x \\ &&&= \int_0^1 \frac1{(x+3)^2} \left ( \ln \frac{x+1}{x+3} + \ln \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_{y=2/3}^{y=3/4} \ln y\, \d y \\ &&&= I + \left [ y \ln y- y\right]_{2/3}^{3/4} \\ &&&= I + \left ( \frac34 \ln \frac34 - \frac34 - \frac23 \ln \frac23 + \frac23 \right) \\ &&&= I + \left ( \frac34 \ln 3 - \frac32 \ln 2- \frac1{12} - \frac23 \ln 2 + \frac23 \ln 3\right) \\ &&&= I + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac{19}{12} \ln 3 -\frac{29}{12}\ln 2 - \frac16 \end{align*} \begin{align*} && K &= \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x \\ &&&= \int_0^1 \frac{1}{(x+3)^2} \; \left ( \ln\left(\frac{x+1}{x+3}\right) - \ln \left ( \frac{x+3}{x+2} \right) \right)\d x \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} - \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= -\frac54 \ln 3 +\frac{23}{12} \ln 2 \end{align*}

---

Solution: \begin{align*} && y &= \frac x {\sqrt{x^2-2x+a}} \\ && y' &= \frac{\sqrt{x^2-2x+a} - \frac{x(x-1)}{\sqrt{x^2-2x+a}}}{x^2-2x+a} \\ &&&= \frac{-x+a}{(x^2-2x+a)^{3/2}} \end{align*} Since the denominator is always positive, the only stationary point is when \(x = a\)