Mathematics

Integrate the rational function: $$\cfrac {1}{x^2-9}$$


SOLUTION
Let $$\displaystyle \frac {1}{(x+3)(x-3)}=\frac {A}{(x+3)}+\frac {B}{(x-3)}$$

$$\Rightarrow 1=A(x-3)+B(x+3)$$

Equating the coefficients of $$x$$ and constant term, we obtain

$$A + B = 0,3A + 3B = 1$$

On solving, we obtain

$$A=-\cfrac {1}{6}$$ and $$B=\cfrac {1}{6}$$

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

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

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

$$\displaystyle =\frac {1}{6}\log \left |\frac {(x-3)}{(x+3)}\right |+C$$
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Subjective Medium Published on 17th 09, 2020
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