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

Author(s):

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: $\int \frac{dx}{\sqrt{2ax+x^2}}= \log \left( x + a + \sqrt{(2ax+x^2)} \right ) + \log ⁡2. \mbox{ (Omit the constant.) }$

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.

Adrian Rice (Randolph-Macon College), "Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Problem 9," Convergence (September 2021)