Question

\[\int\frac{a x^2 + bx + c}{\left( x - a \right) \left( x - b \right) \left( x - c \right)} dx,\text{ where a, b, c are distinct}\]

Answer 1

We have,

\[I = \int\frac{a x^2 + bx + c}{\left( x - a \right) \left( x - b \right) \left( x - c \right)} dx\]

\[\text{Let }\frac{a x^2 + bx + c}{\left( x - a \right) \left( x - b \right) \left( x - c \right)} = \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}\]

\[ \Rightarrow a x^2 + bx + c = A\left( x - b \right) \left( x - c \right) + B \left( x - c \right)\left( x - a \right) + C\left( x - a \right) \left( x - b \right)\]

\[ \Rightarrow a x^2 + bx + c = A\left[ x^2 - \left( b + c \right)x + bc \right] + B\left[ x^2 - \left( c + a \right)x + ca \right] + C\left[ x^2 - \left( a + b \right)x + ab \right]\]

\[ \Rightarrow a x^2 + bx + c = \left( A + B + C \right) x^2 - \left[ A\left( b + c \right) + B\left( c + a \right) + C\left( a + b \right) \right]x + Abc + Bca + Cab\]

Equating the coefficients on both sides, we get

\[a = A + B + C ...............(1)\]

\[b = - \left[ A\left( b + c \right) + B\left( c + a \right) + C\left( a + b \right) \right] ..................(2)\]

\[c = Abc + Bca + Cab .................(3)\]

Solving (1), (2) and (3), we get

\[A = \frac{a^3 + ab + c}{\left( a - b \right)\left( a - c \right)}\]

\[B = \frac{a b^2 + b^2 + c}{\left( b - a \right)\left( b - c \right)}\]

\[C = \frac{a c^2 + bc + c}{\left( c - a \right)\left( c - b \right)}\]

\[ \therefore I = \int\left[ \frac{a^3 + ab + c}{\left( a - b \right)\left( a - c \right)} \times \frac{1}{x - a} + \frac{a b^2 + b^2 + c}{\left( b - a \right)\left( b - c \right)} \times \frac{1}{x - b} + \frac{a c^2 + bc + c}{\left( c - a \right)\left( c - b \right)} \times \frac{1}{x - c} \right] dx\]

\[ = \frac{a^3 + ab + c}{\left( a - b \right)\left( a - c \right)}\log \left| x - a \right| + \frac{a b^2 + b^2 + c}{\left( b - a \right)\left( b - c \right)}\log \left| x - b \right| + \frac{a c^2 + bc + c}{\left( c - a \right)\left( c - b \right)}\log \left| x - c \right| + K\]