Question

express the integrand as a sum of partial fractions and evaluate the integral
int\frac{x + 3}{x^{2}+8x - 9}dx
express the integrand as a sum of partial fractions
\frac{x + 3}{x^{2}+8x - 9}=square
(use integers or fractions for any numbers in the expression )

Explanation:

Step1: Factor the denominator

First, factor \(x^{2}+8x - 9=(x + 9)(x-1)\). Then we write \(\frac{x + 3}{x^{2}+8x - 9}=\frac{x + 3}{(x + 9)(x - 1)}=\frac{A}{x+9}+\frac{B}{x - 1}\).

Step2: Find the values of A and B

Cross - multiply: \(x + 3=A(x - 1)+B(x + 9)\). Let \(x=1\), then \(1+3=A(1 - 1)+B(1 + 9)\), \(4 = 10B\), so \(B=\frac{2}{5}\). Let \(x=-9\), then \(-9 + 3=A(-9 - 1)+B(-9 + 9)\), \(-6=-10A\), so \(A=\frac{3}{5}\). So \(\frac{x + 3}{x^{2}+8x - 9}=\frac{3}{5(x + 9)}+\frac{2}{5(x - 1)}\).

Step3: Evaluate the integral

\(\int\frac{x + 3}{x^{2}+8x - 9}dx=\int(\frac{3}{5(x + 9)}+\frac{2}{5(x - 1)})dx=\frac{3}{5}\int\frac{1}{x + 9}dx+\frac{2}{5}\int\frac{1}{x - 1}dx\). Using the formula \(\int\frac{1}{u}du=\ln|u|+C\), we get \(\frac{3}{5}\ln|x + 9|+\frac{2}{5}\ln|x - 1|+C\).

Answer:

\(\frac{x + 3}{x^{2}+8x - 9}=\frac{3}{5(x + 9)}+\frac{2}{5(x - 1)}\), \(\int\frac{x + 3}{x^{2}+8x - 9}dx=\frac{3}{5}\ln|x + 9|+\frac{2}{5}\ln|x - 1|+C\)