Home page for accesible maths 1 1 Further Integration

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

1.15 Integrating the partial fractions

Integrating the rational function f(x)g(x){f(x)\over g(x)} is therefore reduced to integrating a sum of a polynomial and some expressions of the form

(i)γ(x-a)ror(ii)Hx+K(x2+2bx+c)r.(i)\quad{{\gamma}\over{(x-a)^{r}}}\quad{\hbox{or}}\quad(ii)\quad{{Hx+K}\over{(% x^{2}+2bx+c)^{r}}}.

Case (i) is straightforward to integrate as in Chapter 5 of MATH101: substituting u=x-au=x-a we have dudx=1{du\over{dx}}=1 and hence

γx-adx=γduu=γlog|u|+c=γlog|x-a|+c,\int{\gamma\over{x-a}}dx=\int{\gamma\,du\over u}=\gamma\log|u|+c=\gamma\log|x-% a|+c,
γ(x-a)rdx=γurdu=-γ(r-1)ur-1+c=-γ(r-1)(x-a)r-1+c.\int{\gamma\over{(x-a)^{r}}}dx=\int{\gamma\over u^{r}}du=-{\gamma\over{(r-1)u^% {r-1}}}+c=-{\gamma\over{(r-1)(x-a)^{r-1}}}+c.