express the integrand as a sum of partial fractions and evaluate the integral.\n int\frac{x + 3}{x^{2}+5x…

express the integrand as a sum of partial fractions and evaluate the integral.\n int\frac{x + 3}{x^{2}+5x - 6}dx\nx^{2}+5x - 6=x^{2}+6 x^{2}-1\n int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{\frac{3}{7}}{x + 6}+\frac{\frac{4}{7}}{x - 1}\rightdx\n int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{}{x + 3}+\frac{}{x - 3}\rightdx\nevaluate the integral.\n int\frac{x + 3}{x^{2}+5x - 6}dx=square\n(use parentheses to clearly denote the argument of each function.)

express the integrand as a sum of partial fractions and evaluate the integral.\n int\frac{x + 3}{x^{2}+5x - 6}dx\nx^{2}+5x - 6=x^{2}+6 x^{2}-1\n int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{\frac{3}{7}}{x + 6}+\frac{\frac{4}{7}}{x - 1}\rightdx\n int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{}{x + 3}+\frac{}{x - 3}\rightdx\nevaluate the integral.\n int\frac{x + 3}{x^{2}+5x - 6}dx=square\n(use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Factor the denominator

First, factor (x^{2}+5x - 6=(x + 6)(x - 1)). Then we write (\frac{x + 3}{x^{2}+5x - 6}=\frac{x + 3}{(x + 6)(x - 1)}=\frac{A}{x+6}+\frac{B}{x - 1}). Cross - multiply gives (x + 3=A(x - 1)+B(x + 6)). Let (x = 1), then (1+3=A(1 - 1)+B(1 + 6)), so (4 = 7B), (B=\frac{4}{7}). Let (x=-6), then (-6 + 3=A(-6 - 1)+B(-6 + 6)), so (-3=-7A), (A=\frac{3}{7}). So (\frac{x + 3}{x^{2}+5x - 6}=\frac{3/7}{x + 6}+\frac{4/7}{x - 1}).

Step2: Integrate the partial - fraction decomposition

(\int\frac{x + 3}{x^{2}+5x - 6}dx=\int\left(\frac{3/7}{x + 6}+\frac{4/7}{x - 1}\right)dx=\frac{3}{7}\int\frac{1}{x + 6}dx+\frac{4}{7}\int\frac{1}{x - 1}dx). Using the formula (\int\frac{1}{u}du=\ln|u|+C), we get (\frac{3}{7}\ln|x + 6|+\frac{4}{7}\ln|x - 1|+C).

Answer:

(\frac{3}{7}\ln|x + 6|+\frac{4}{7}\ln|x - 1|+C)