Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Problem 9

Adrian Rice (Randolph-Macon College)

The integral \[\int \frac{dx}{\sqrt{A+Bx+Cx^2}}\] is more sophisticated still. Try to prove for yourself that, if \(C>0\), \[\int \frac{dx}{\sqrt{A+Bx+Cx^2}}= \frac{1}{\sqrt{C}} \log\left ( 2Cx+B+\sqrt{4C(A+Bx+Cx^2)} \right ).\] In his Calculus, letting \(A=0\), \(B=2a\), and \(C=1\), De Morgan deduced that, as he put it: \[\int \frac{dx}{\sqrt{2ax+x^2}}= \log \left( x + a + \sqrt{(2ax+x^2)} \right ) + \log ⁡2. \mbox{ (Omit the constant.) } \]

Another excerpt from page 116 of De Morgan's calculus textbook.

Figure 11. Part of page 116 from De Morgan’s Differential and Integral Calculus.

But when Lovelace tried to derive this result from first principles, she wrote [LB 170, 15 Aug. [1841], f. 115v]:

I cannot make it anything but \[\int \frac{dx}{\sqrt{2ax+x^2}} = \log \left ( x + 2a + \sqrt{(2ax+x^2)} \right )\] or else \[ = \log \left ( \frac{x}{2} + a + \frac{\sqrt{2ax+x^2}}{2} \right )+ \log 2⁡  \] \(\ldots\)  and I begin to suspect the book.

Her approach was straightforward. Setting \(2ax+x^2=(2a+x)x=y^2\), she derived the differential equation \((2a+x)dx=ydy\), from which she formed the integral \[\int  \frac{dx}{y}=\int \frac{dy}{2a+x}  .\]

Then, by analogy with the ‘fact’ that \(\frac{dx+dy}{x+y}=\frac{dx}{y}\), she obtained \[ \begin{array}{lcl} \int \frac{dx}{y} &=&  \int \frac{d(2a+x)+dy}{(2a+x)+y}\\ &=&\int \frac{d(2a+x+y)}{2a+x+y}\\ &=&\log⁡(2a+x+y)\\ &=&\log⁡(2a+x+\sqrt{2ax+x^2})\\ &=& \log⁡\left ( \frac{x}{2} + a + \frac{\sqrt{2ax+x^2}}{2}\right )+\log ⁡2.\end{array}\]

In addition to her erroneous assumption that \[\frac{dx+dy}{x+y}=\frac{dx}{y}\]

De Morgan was able to spot that, given \(y^2=(2a+x)x\), she had forgotten to apply the product rule, so that her differential equation should have been \[ydy=\frac{1}{2} xdx + \frac{1}{2} (2a+x)dx \] or   \[ydy=(x+a)dx.\,\,\,\,\,               \mbox{[2]}\] After correcting these errors, Lovelace wrote in a subsequent letter [LB 170, 21 Aug. [1841], f. 121v], that

we arrive then in my corrected paper, at \[\begin{array}{rcl}\int \frac{dx}{\sqrt{2ax+x^2}} &=& \log \left( x + a + \sqrt{(2ax+x^2)} \right )\\ &=&\log⁡\left( \frac{x}{2} + \frac{a}{2} + \frac{\sqrt{2ax+x^2}}{2} \right )+\log ⁡2.\end{array}\]

Can you derive Lovelace’s final result from equation [2]?

In his acknowledgement, De Morgan observed that Lovelace’s answer agreed with his result in all respects ‘but the log 2, which being a Constant, matters nothing’ [LB 170, 21 Aug. [1841], f. 121v]. Explain why his and Lovelace’s results are equivalent.

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