Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 445: 12

$$\int\frac{2x+1}{x^2-7x+12}dx=9\ln|x-4|-7\ln|x-3|+C$$

Work Step by Step

$$A=\int\frac{2x+1}{x^2-7x+12}dx$$ 1) Express the integrand as a sum of partial fractions: $$\frac{2x+1}{x^2-7x+12}=\frac{2x+1}{(x-3)(x-4)}=\frac{A}{x-3}+\frac{B}{x-4}$$ Clear fractions: $$2x+1=Ax-4A+Bx-3B$$ $$2x+1=(A+B)x+(-4A-3B)$$ Equating coefficients of corresponding powers of $x$, we get - $A+B=2$ - $-4A-3B=1$ Calculation gives us $A=-7$ and $B=9$. Therefore, $$\frac{2x+1}{x^2-7x+12}=\frac{9}{x-4}-\frac{7}{x-3}$$ 2) Evaluate the integral: $$A=9\int\frac{dx}{x-4}-7\int\frac{dx}{x-3}$$ $$A=9\ln|x-4|-7\ln|x-3|+C$$

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