Integrate the rational function: \(\frac{2x-3}{(x^2-1)(2x+3)}\)
Solution and Explanation
\(\frac{2x-3}{(x^2-1)(2x+3)}\)= \(\frac{2x-3}{(x+1)(x-1)(2x+3)}\)
Let \(\frac{2x-3}{(x+1)(x-1)(2x+3)}= \frac{A}{(x+1)}\frac{B}{(x-1)}+\frac{C}{(2x+3)}\)
\(\Rightarrow\) (2x-3) = A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)
\(\Rightarrow\) (2x-3) = A(2x2+x-3)+B(2x2+5x+3)+C(x2-1)
\(\Rightarrow\) (2x-3) = A(2A+2B+C)x2+(A+5B)x+(-3A+3B-C)
Equating the coefficients of x2 and x, we obtain
B = -\(\frac{1}{10}\), A = \(\frac{5}{2}\), and C = -\(\frac{24}{5}\)
∴ \(\frac{2x-3}{(x+1)(x-1)(2x+3)}= \frac{5}{2(x+1)}\frac{1}{10(x-1)}+\frac{24}{5(2x+3)}\)
\(\Rightarrow\int\frac{2x-3}{(x+1)(x-1)(2x+3)}dx= \frac{5}{2}\int\frac{1}{(x+1)}dx-\frac{1}{10}\int\frac{1}{x-1}dx-\frac{24}{5}\int\frac{1}{(2x+3)}dx\)
= \(\frac{5}{2}\log\mid x+1\mid-\frac{1}{10}\log \mid x-1\mid-\frac{24}{5*2}\log\mid 2x+3\mid\)
=\(\frac{5}{2}\log\mid x+1\mid-\frac{1}{10}\log\mid x-1\mid-\frac{12}{5}\log\mid 2x+3 \mid+C\)
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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,
