Page:Calculus Made Easy.pdf/220

the problem how to integrate $x^{-1}\,dx$. Indeed it should be frankly admitted that this is one of the curious features of the integral calculus:–that you can’t integrate anything before the reverse process of differentiating something else has yielded that expression which you want to integrate. No one, even to-day, is able to find the general integral of the expression,

${\frac {dy}{dx}}=a^{-x^{2}}$,

because $a-x^{2}$ has never yet been found to result from differentiating anything else.

Another simple case.

Find $\int (x+1)(x+2)\,dx$.

On looking at the function to be integrated, you remark that it is the product of two different functions of $x$. You could, you think, integrate $(x+1)\,dx$ by itself, or $(x+2)\,dx$ by itself. Of course you could. But what to do with a product? None of the differentiations you have learned have yielded you for the differential coefficient a product like this. Failing such, the simplest thing is to multiply up the two functions, and then integrate. This gives us

$\int (x^{2}+3x+2)\,dx$.

And this is the same as

$\int x^{2}\,dx+\int 3x\,dx+\int 2\,dx$.

And performing the integrations, we get

${\tfrac {1}{3}}x^{3}+{\tfrac {3}{2}}x^{2}+2x+C$.