# Calculus Made Easy - Chapter 20, Exercise 19.14

This book is one of my favorites. While I eventually learned "proper" calculus at university, it was Sylvanus P. Thompson's slim text about infinitesimals that got me started. From the prologue:

Being myself a remarkably stupid fellow, I have had to unteach myself the difficulties, and now beg to present to my fellow fools the parts that are not hard.

I can relate. In that spirit, let us consider the solution to:

$$ \int \frac{1}{x \sqrt{a - bx^2}} dx $$

This is from the original book; Martin Gardner's updated edition substitutes a simpler exercise. And understandably so — this is a doozy of a problem for an autodidact of modest talent.

Fortunately, we are not on our own. Flipping to the solution in frustration, we find this hint from the author:

Let \(\dfrac{1}{x} = v\); then, in the result, let \(\sqrt{v^2 - \dfrac{b}{a}} = v - u\).

This will ultimately prove to be an exercise in handwriting and concentration as much as anything.

Obviously, we are meant to perform integration by substitution here, rewriting the equation to be in terms of \(v\) rather than \(x\). We will also need to know how to substitute away the \(dx\) term, so we calculate:

$$ x = \frac{1}{v} $$ $$ \frac{dx}{dv} = \frac{-1}{v^2} $$ $$ dx = \frac{-1}{v^2} dv $$

Performing the substitution:

$$ \int \frac{1}{x \sqrt{a - bx^2}} dx = \int \frac{1}{x} \frac{1}{\sqrt{a - bx^2}} dx = \int v \frac{1}{\sqrt{a - bx^2}} \frac{-1}{v^2} dv $$

$$ = \int \frac{-v}{v^2\sqrt{a - bx^2}} dv = \int \frac{-1}{v\sqrt{a - bx^2}} dv $$

Recall that \( x = \frac{1}{v} \). This implies that \( x^2 = \frac{1}{v^2} \), which gives us another opportunity to substitute and simplify:

$$ \int \frac{-1}{v\sqrt{a - bx^2}} dv = \int \frac{-1}{v\sqrt{a - \frac{b}{v^2}}} dv = \int \frac{-1}{v\sqrt{a\frac{v^2}{v^2} - \frac{b}{v^2}}} dv $$

$$ = \int \frac{-1}{v\sqrt{\frac{av^2-b}{v^2}}} dv = \int \frac{-1}{v\frac{\sqrt{av^2-b}}{\sqrt{v^2}}} dv = \int \frac{-1}{v\frac{\sqrt{av^2-b}}{v}} dv $$

$$ = \int \frac{-1}{\sqrt{av^2-b}} dv $$

What comes next would be a mystery to this particular fool were it not for the second part of the author's hint:

$$ v - u = \sqrt{v^2 - \frac{b}{a}} $$

With a little work, we can modify the denominator of our equation to be substitutable for \( v - u \):

$$ \int \frac{-1}{\sqrt{av^2-b}} dv = \int \frac{-1}{\sqrt{av^2-b}} \frac{\sqrt{\frac{1}{a}}}{\sqrt{\frac{1}{a}}} dv = \frac{-1}{\sqrt{a}} \int \frac{1}{\sqrt{v^2-\frac{b}{a}}} dv $$

Integration is all about guessing and testing. At this point, we can guess that we would like to further simplify our equation by expressing it in terms of \( u \). It may not be immediately obvious that this will work, but we can give it a shot. To do that, we will need to replace \( dv \) with \( du \), which we accomplish by first finding \( \frac{du}{dv} \):

$$ u = v - \sqrt{v^2 - \frac{b}{a}} = v - (v^2 - \frac{b}{a})^\frac{1}{2} $$

$$ \frac{du}{dv} = 1 - \frac{1}{2}(v^2 - \frac{b}{a})^\frac{-1}{2} (2v) = 1 - \frac{2v}{2\sqrt{v^2 - \frac{b}{a}}} $$

$$ \frac{du}{dv} = 1 - \frac{v}{\sqrt{v^2 - \frac{b}{a}}} $$

We can further simplify this using the hint that \(\sqrt{v^2 - \dfrac{b}{a}} = v - u\):

$$ \frac{du}{dv} = 1 - \frac{v}{\sqrt{v^2 - \frac{b}{a}}} = 1 - \frac{v}{v - u} = \frac{v - u}{v - u} - \frac{v}{v - u} $$

$$ \frac{du}{dv} = \frac{(v - u) - v}{v - u} = \frac{-u}{v - u} $$

Then we rearrange for \(dv\) in terms of \(du\):

$$ dv = \frac{v-u}{-u} du $$

Now to see if we can further simplify our problem:

$$ \frac{-1}{\sqrt{a}} \int \frac{1}{\sqrt{v^2-\frac{b}{a}}} dv = \frac{-1}{\sqrt{a}} \int \frac{1}{v - u} \frac{v-u}{-u} du = \frac{-1}{\sqrt{a}} \int \frac{1}{ -u} du $$

$$ = \frac{1}{\sqrt{a}} \int \frac{1}{u} du $$

Our guess paid off! We know from the rules of differentiation that \(\frac{d \ln{x}}{dx} = \frac{1}{x} \), so we can easily find the antiderivative of our problem:

$$ \frac{1}{\sqrt{a}} \int \frac{1}{u} du = \frac{1}{\sqrt{a}} \ln{u} $$

Now we work backwards to our solution. Recall that:

$$ u = v - \sqrt{v^2 - \frac{b}{a}} $$

Therefore:

$$ \frac{1}{\sqrt{a}} \ln{u} = \frac{1}{\sqrt{a}} \ln{(v - \sqrt{v^2 - \frac{b}{a}})} $$

Remembering that \(v = \frac{1}{x}\) gets us the rest of the way there:

$$ \frac{1}{\sqrt{a}} \ln{(v - \sqrt{v^2 - \frac{b}{a}})} = \frac{1}{\sqrt{a}} \ln{(\frac{1}{x} - \sqrt{\frac{1}{x^2} - \frac{b}{a}})} $$

$$ = \frac{1}{\sqrt{a}} \ln{(\frac{1}{x} - \sqrt{\frac{1}{x^2}\frac{a}{a} - \frac{b}{a}\frac{x^2}{x^2}})} = \frac{1}{\sqrt{a}} \ln{(\frac{1}{x} - \sqrt{\frac{a - bx^2}{ax^2}})} $$

$$ = \frac{1}{\sqrt{a}} \ln{(\frac{1}{x} - \frac{\sqrt{a - bx^2}}{\sqrt{ax^2}})} = \frac{1}{\sqrt{a}} \ln{(\frac{1}{x} - \frac{\sqrt{a - bx^2}}{x\sqrt{a}})} $$

$$ = \frac{1}{\sqrt{a}} \ln{(\frac{1}{x}\frac{\sqrt{a}}{\sqrt{a}} - \frac{\sqrt{a - bx^2}}{x\sqrt{a}})} = \frac{1}{\sqrt{a}} \ln{(\frac{\sqrt{a}-\sqrt{a - bx^2}}{x\sqrt{a}})} $$

And because it is good to have a little decorum about ourselves, we include the constant of integration to complete our answer:

$$\int \frac{1}{x \sqrt{a - bx^2}} dx = \frac{1}{\sqrt{a}} \ln{(\frac{\sqrt{a}-\sqrt{a - bx^2}}{x\sqrt{a}})} + C$$

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