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Q 3  Integrate the rational functions \frac{3x -1}{( x-1)(x-2)(x-3)}

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Given function \frac{3x -1}{( x-1)(x-2)(x-3)}

Partial function of this function:

\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)

Now, substituting x=1,2,\ and\ 3 respectively in equation (1), we get

A =1,\ B=-5,\ and\ C=4

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

That implies \int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)} \right \}dx

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

 

 

Posted by

Divya Prakash Singh

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