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

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Since  $y^2=2ax+x^2=(2a+x)x$ in Lovelace's approach to Problem 9,  using the product rule we have   $ydy=(x+a)dx \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [2]$ so that $(x+a)dy+ydy=(x+a)(dx+dy)$ or $(x+a+y)dy=(x+a)(dx+da+dy)$ and thus $\frac{dy}{x+a} = \frac{d(x+a+y)}{x+a+y}.$ Now since, by [2], $\frac{dy}{x+a}=\frac{dx}{y}$ this means that $\frac{dx}{y}=\frac{d(x+a+y)}{x+a+y}.$ So, recalling that $y^2=2ax+x^2$, we have $\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(x+a+\sqrt{2ax+x^2}) = \log\left (\frac {x}{2} +\frac{a}{2} + \frac{\sqrt{2ax+x^2}}{2} \right )+ \log ⁡2$ which is Lovelace’s result.
Now recall that De Morgan had obtained $\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(x+a+\sqrt{2ax+x^2}) + \log ⁡2.$ Strictly speaking, his result should read $\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(2x+2a+2\sqrt{2ax+x^2}) +C$ and Lovelace’s result should be $\int \frac{dx}{\sqrt{2ax+x^2}} = \log⁡(x + a +\sqrt{2ax+x^2}) +C'.$ Thus, letting Lovelace’s constant of integration $C'=C+\log ⁡2$ reveals that the two results are equivalent and also explains De Morgan’s remark that the $\log 2$, ‘being a Constant, matters nothing.’