Evaluate the following integral:

Denominator is factorised, so let separate the fraction through partial fraction, hence let

$\frac{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})}=\frac{\mathrm{A}}{(\mathrm{x}-\mathrm{a})}+\frac{\mathrm{B}}{\mathrm{x}-\mathrm{b}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{c}} \ldots$(i)

$\Rightarrow \frac{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})}$

$=\frac{\mathrm{A}(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})+\mathrm{B}(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{c})+\mathrm{C}(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})}$

$\Rightarrow a x^{2}+b x+c=A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b) \ldots \ldots(i i)$

We need to solve for A, B and C. One way to do this is to pick values for $x$ which will cancel each variable.

Put $x=a$ in the above equation, we get

$\Rightarrow a(a)^{2}+b(a)+c=A(a-b)(a-c)+B(a-a)(a-c)+C(a-a)(a-b)$

$\Rightarrow a^{3}+a b+c=(a-b)(a-c) A+0+0$

$\Rightarrow A=\frac{a^{3}+a b+c}{(a-b)(a-c)}$

Now put $x=b$ in equation (ii), we get

$\Rightarrow a(b)^{2}+b(b)+c=A(b-b)(b-c)+B(b-a)(b-c)+C(b-a)(b-b)$

$\Rightarrow a b^{2}+b^{2}+c=0+(b-a)(b-c) B+0$

$\Rightarrow \mathrm{B}=\frac{\mathrm{a}^{3}+\mathrm{ab}+\mathrm{c}}{(\mathrm{a}-\mathrm{b})(\mathrm{a}-\mathrm{c})}$

Now put $x=c$ in equation (ii), we get

$\Rightarrow a(c)^{2}+b(c)+c$

$=A(c-b)(c-c)+B(c-a)(c-c)+C(c-a)(c-b)$

$\Rightarrow \mathrm{ac}^{2}+\mathrm{bc}+\mathrm{c}=0+0+(\mathrm{c}-\mathrm{a})(\mathrm{c}-\mathrm{b}) \mathrm{C}$

$\Rightarrow C=\frac{a c^{2}+b c+c}{(c-a)(c-b)}$

We put the values of $A, B$, and $C$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int\left[\frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)}\right] d x$

$\Rightarrow \int\left[\frac{\mathrm{A}}{(\mathrm{x}-\mathrm{a})}+\frac{\mathrm{B}}{\mathrm{x}-\mathrm{b}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{c}}\right] \mathrm{dx}$

$\Rightarrow \int\left[\frac{\frac{a^{3}+a b+c}{(a-b)(a-c)}}{(x-a)}+\frac{\frac{a^{3}+a b+c}{(a-b)(a-c)}}{x-b}+\frac{\frac{a c^{2}+b c+c}{(c-a)(c-b)}}{x-c}\right] d x$

Split up the integral,

$\Rightarrow \frac{a^{3}+a b+c}{(a-b)(a-c)} \int \frac{1}{x-a} d x+\frac{a^{3}+a b+c}{(a-b)(a-c)} \int\left[\frac{1}{x-b}\right] d x$

$+\frac{a c^{2}+b c+c}{(c-a)(c-b)} \int\left[\frac{1}{x-c}\right] d x$

Let substitute

$u=x-a \Rightarrow d u=d x$

$y=x-b \Rightarrow d y=d x$ and

$z=x-c \Rightarrow d z=d x$, so the above equation becomes,

$\Rightarrow \frac{a^{3}+a b+c}{(a-b)(a-c)} \int \frac{1}{u} d u+\frac{a^{3}+a b+c}{(a-b)(a-c)} \int\left[\frac{1}{y}\right] d y+\frac{a c^{2}+b c+c}{(c-a)(c-b)} \int\left[\frac{1}{z}\right] d z$

On integrating we get

$\Rightarrow \frac{a^{3}+a b+c}{(a-b)(a-c)} \log |u|+\frac{a^{3}+a b+c}{(a-b)(a-c)} \log |y|+\frac{a c^{2}+b c+c}{(c-a)(c-b)} \log |z|+C$

Substituting back, we get

$\Rightarrow \frac{a^{3}+a b+c}{(a-b)(a-c)} \log |x-a|+\frac{a^{3}+a b+c}{(a-b)(a-c)} \log |x-b|$

$+\frac{a c^{2}+b c+c}{(c-a)(c-b)} \log |x-c|+c$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)} d x$

$=\frac{a^{3}+a b+c}{(a-b)(a-c)} \log |x-a|+\frac{a^{3}+a b+c}{(a-b)(a-c)} \log |x-b|$

$+\frac{a c^{2}+b c+c}{(c-a)(c-b)} \log |x-c|+c$