If `g(x)=(f(x))/((x-a)(x-b)(x-c)),w h e r ef(x)` is a polynomial of degree `<3` , then `intg(x)dx=|1af(a)log|x-a|1bf(b)log|x-b|1cf(c)log|x-c||-:|1a a^2 1 bb^2 1 cc^2|+k` `(dg(x))/(dx)=|1af(a)log""(x-a)^2 1bf(b)"log"(x-b)^2 1cf(c)"log"(x-b)^2|-:|a^2a1b^2b1c^2c1|` `(dg(x))/(dx)=|1af(a)log""(x-a)^(-2)1bf(b)"log"(x-b)^(-2)1cf(c)"log"(x-b)^(-2)|-:|1a a^2 1bb^2 1cc^2|` `intg(x)dx=|1af(a)"log"|x-a|1bf(b)"log"|x-b|1cf(c)"log"|x-c||-:|a^2a1b^2b1c^2c1|+k`

Explanation

By partial fractions, we have
`g(x)= g(x)=(f(a))/((x-a)(a-b)(a-c))+(f(b))/((b-a)(x-b)(b-c))+(f(c))/((b-a)(c-b)(x-c))`
`=(1)/((a-b)(b-c)(c-a))`
`xx[(f(a)(c-b))/((x-a))+(f(b)(a-c))/((x-b))+(f(c)(b-a))/((x-c))]`
`=|{:(1,a,f(a)//(x-a)),(1,b,f( b)//(x-b)),(1,c,f(c)//(x-c)):}|div|{:(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2)):}|`
`therefore" "(dg(x))/(dx)=|{:(1,a,-f(a)//(x-a)^(-2)),(1,b,-f( b)//(x-b)^(-2)),(1,c,-f(c)//(x-c)^(-2)):}|div|{:(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2)):}|`
`=|{:(1,a,f(a)//(x-a)^(-2)),(1,b,f(b)//(x-b)^(-2)),(1,c,f(c)//(x-c)^(-2)):}|div|{:(a^(2),a,1),(b^(2),b,1),(c^(2),c,1):}|`