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Q 2  Integrate the rational functions \frac{1}{x^2 -9 }

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Given function \frac{1}{x^2 -9 }

The partial function of this function:

\frac{1}{(x+3)(x-3)}= \frac{A}{(x+3)}+\frac{B}{(x-3)}

1 = A(x-3)+B(x+3)

Now, equating the coefficients of x and constant term, we obtain

A+B =1

-3A+3B =1

On solving, we get

A=-\frac{1}{6}\ and\ B =\frac{1}{6}

\frac{1}{(x+3)(x-3)}= \frac{-1}{6(x+3)} +\frac{1}{6(x-3)}

\int \frac{1}{(x^2-9)}dx = \int \left ( \frac{-1}{6(x+3)}+\frac{1}{6(x-3)} \right )dx

=-\frac{1}{6}\log|x+3| +\frac{1}{6}\log|x-3| +C

= \frac{1}{6}\log\left | \frac{x-3}{x+3} \right |+C

 

 

Posted by

Divya Prakash Singh

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