Page:Calculus Made Easy.pdf/250

Let $x+1=u,\quad dx=du$;

then the integral becomes $\int {\dfrac {du}{u^{2}+({\sqrt {2}})^{2}}}$; but ${\dfrac {du}{u^{2}+a^{2}}}$ is the result of differentiating $u={\dfrac {1}{a}}\arctan {\dfrac {u}{a}}$.

Hence one has finally ${\dfrac {1}{\sqrt {2}}}\arctan {\dfrac {x+1}{\sqrt {2}}}$ for the value of the given integral.

Formulæ of Reduction are special forms applicable chiefly to binomial and trigonometrical expressions that have to be integrated, and have to be reduced into some form of which the integral is known.

Rationalization, and Factorization of Denominator are dodges applicable in special cases, but they do not admit of any short or general explanation. Much practice is needed to become familiar with these preparatory processes.

The following example shows how the process of splitting into partial fractions, which we learned in Chap. XIII., p. 122, here, can be made use of in integration.

Take again $\int {\dfrac {dx}{x^{2}+2x+3}}$; if we split ${\dfrac {1}{x^{2}+2x+3}}$ into partial fractions, this becomes (see p. 232):

${\dfrac {1}{2{\sqrt {-2}}}}\left[\int {\dfrac {dx}{x+1-{\sqrt {-2}}}}-\int {\dfrac {dx}{x+1+{\sqrt {-2}}}}\right]$
$={\dfrac {1}{2{\sqrt {-2}}}}\log _{\epsilon }{\dfrac {x+1-{\sqrt {-2}}}{x+1+{\sqrt {-2}}}}$.

Notice that the same integral can be expressed