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