Integrate the rational function: \(\frac{1}{x^2-9}\)
Solution and Explanation
Let \(\frac{1}{(x+3)(x-3)} = \frac{A}{(x+3)}+\frac{B}{(x-3)}\)
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 =- \(\frac{1}{6}\)and B = \(\frac{1}{6}\)
∴ \(\frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\)
\(\Rightarrow \int\frac{1}{(xx^2-9)}dx=\int\bigg(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\bigg)dx\)
= \(-\frac{1}{6}\log\mid x+3 \mid +\frac{1}{6}\log \mid x-3 \mid+C\)
\(\frac{1}{6}\log \mid\frac{(x-3)}{(x+3)} \mid+C\)
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Concepts Used:
Integration by Partial Fractions
The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.
For examples,
