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Question

Mathematics Question on integral

Integrate the rational function: 3x−1(x−1)(x−2)(x−3)\frac{3x-1}{(x-1)(x-2)(x-3)}

Answer

Let 3x−1(x−1)(x−2)(x−3)=A(x−1)+B(x−2)+C(x−3)\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}

3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) ...(1)

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

A = 1, B = −5, and C = 4

∴ 3x−1(x−1)(x−2)(x−3)=1(x−1)−5(x−2)+4(x−3)\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}

⇒∫3x−1(x−1)(x−2)(x−3)dx=∫1(x−1)−5(x−2)+4(x−3)dx\Rightarrow \int\frac{3x-1}{(x-1)(x-2)(x-3)}dx =\int \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}dx

= log|x-1|-5log|x-2|+4log|x-3|+C