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\nexpress the integrand as a sum of partial fractions. select the correct choice below and fill (use integers or fractions for any numbers in the expression.)\na. (int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{square}{x^{2}+6}+\frac{square}{x^{2}-1}\rightdx)\nb. (int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{square}{x + 6}+\frac{square}{x - 1}\rightdx)\nc. (int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{square}{x + 3}+\frac{square}{x - 3}\rightdx)

express the integrand as a sum of partial fractions and evaluate the integral.\n int\frac{x + 3}{x^{2}+5x - 6}dx\nexpress the integrand as a sum of partial fractions. select the correct choice below and fill (use integers or fractions for any numbers in the expression.)\na. (int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{square}{x^{2}+6}+\frac{square}{x^{2}-1}\rightdx)\nb. (int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{square}{x + 6}+\frac{square}{x - 1}\rightdx)\nc. (int\frac{x + 3}{x^{2}+5x - 6}dx=intleft\frac{square}{x + 3}+\frac{square}{x - 3}\rightdx)

Answer

Explanation:

Step1: Factor the denominator

First, factor (x^{2}+5x - 6=(x + 6)(x-1)).

Step2: Set up partial - fraction decomposition

Let (\frac{x + 3}{x^{2}+5x - 6}=\frac{A}{x + 6}+\frac{B}{x - 1}). Then (x + 3=A(x - 1)+B(x + 6)).

Step3: Find the values of A and B

If (x=1), then (1 + 3=A(1 - 1)+B(1 + 6)), so (4 = 7B), and (B=\frac{4}{7}). If (x=-6), then (-6 + 3=A(-6 - 1)+B(-6 + 6)), so (-3=-7A), and (A=\frac{3}{7}). So (\frac{x + 3}{x^{2}+5x - 6}=\frac{3/7}{x + 6}+\frac{4/7}{x - 1}), and (\int\frac{x + 3}{x^{2}+5x - 6}dx=\int\left[\frac{3/7}{x + 6}+\frac{4/7}{x - 1}\right]dx).

Answer:

B. (\int\frac{x + 3}{x^{2}+5x - 6}dx=\int\left[\frac{3/7}{x + 6}+\frac{4/7}{x - 1}\right]dx)