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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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