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2014 Paper 2 Q4
D: 1600.0 B: 1500.0
integration substitution u=1/x substitution definite integral logarithm arctan symmetry argument parameter differentiation

  1. By using the substitution \(u=1/x\), show that for \(b>0\) \[ \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1)} \d x =0 \,. \]
  2. By using the substitution \(u=1/x\), show that for \(b>0\), \[ \int_{1/b}^b \frac{\arctan x}{x} \d x = \frac{\pi \ln b} 2\,. \]
  3. By using the result \( \displaystyle \int_0^\infty \frac 1 {a^2+x^2} \d x = \frac {\pi}{2 a} \) (where \(a > 0\)),and a substitution of the form \(u=k/x\), for suitable \(k\), show that \[ \int_0^\infty \frac 1 {(a^2+x^2)^2} \d x = \frac {\pi}{4a^3 } \, \ \ \ \ \ \ (a > 0). \]

Solution:

  1. \begin{align*} && I &= \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1} \d x \\ u = 1/x, \d u = -1/x^2 \d x: &&&= \int_{u=b}^{u=1/b} \frac{1/u \ln(1/u)}{(a^2+u^{-2})(a^2u^{-2}+1)} (- \frac{1}{u^2}) \d u \\ &&&= \int_{1/b}^b \frac{-u\ln u}{(a^2u^2+1)(a^2+u^2)} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
  2. \(\,\) \begin{align*} && I &= \int_{1/b}^b \frac{\arctan x}{x} \d x \\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=b}^{u=1/b} \frac{\arctan \frac1u}{\frac1u} \frac{-1}{u^2} \d u \\ &&&= \int_{1/b}^b \frac{\arctan \frac1u}{u} \d u \\ \Rightarrow && 2I &= \int_{1/b}^b \frac{\arctan x + \arctan \frac1x}{x} \d x \\ &&&= \int_{1/b}^b \frac{\frac{\pi}2}{x} \d x \\ &&&= \pi \ln b \\ \Rightarrow && I &= \frac{\pi}{2} \ln b \end{align*}
  3. \(\,\) \begin{align*} && I_a &= \int_0^{\infty} \frac{1}{(a^2+x^2)^2} \d x \\ u = a/x, \d x = -a/u^2 \d u:&&&= \int_{u=0}^{u=\infty} \frac{1}{\left (a^2+\frac{a^2}{u^2} \right)^2} \frac{a}{u^2} \d u \\ &&&= \frac1{a^3}\int_0^{\infty} \frac{1}{(u+1/u)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2+1-1}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{1}{(u^2+1)} - \frac{1}{(u^2+1)^2} \d u \\ &&&= \frac1{a^3} \frac{\pi}{2} - \frac{1}{a^3} I_1 \\ \Rightarrow && 2I_1 &= \frac{\pi}{2} \\ \Rightarrow && I_1 &= \frac{\pi}{4} \\ \Rightarrow && I_a &= \frac{\pi}{4a^3} \end{align*}

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2001 Paper 1 Q5
D: 1500.0 B: 1544.5
integration differentiation under the integral sign rational function parameter differentiation conjecture definite integral

Show that (for \(t>0\))

  1. \[ \int_0^1 \frac1{(1+tx)^2} \d x = \frac1{(1+t)} \]
  2. \[ \int_0^1 \frac{-2x}{(1+tx)^3} \d x = -\frac1{(1+t)^2} \]
Noting that the right hand side of (ii) is the derivative of the right hand side of (i), conjecture the value of \[ \int_0^1 \frac{6x^2}{(1+x)^{4}} \d x \;. \] (You need not verify your conjecture.)

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.

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