express the integrand as a sum of partial fractions and evaluate the integrals. $int\frac{x +…

Explanation:

Step1: Factor the denominator

First, factor out the common factor from the denominator $2x^{3}-18x = 2x(x^{2}-9)=2x(x - 3)(x + 3)$. Then, assume $\frac{x + 7}{2x^{3}-18x}=\frac{A}{2x}+\frac{B}{x - 3}+\frac{C}{x + 3}$. Cross - multiply to get $x + 7=A(x^{2}-9)+B(2x)(x + 3)+C(2x)(x - 3)$.

Step2: Find the values of A, B, and C

Let $x = 0$, then $0+7=A(0 - 9)+B\times0+C\times0$, so $A=-\frac{7}{9}$. Let $x = 3$, then $3 + 7=A\times0+B(2\times3)(3 + 3)+C\times0$, $10 = 36B$, so $B=\frac{5}{18}$. Let $x=-3$, then $-3 + 7=A\times0+B\times0+C(2\times(-3))(-3 - 3)$, $4 = 36C$, so $C=\frac{1}{9}$. So $\frac{x + 7}{2x^{3}-18x}=-\frac{7}{18x}+\frac{5}{18(x - 3)}+\frac{1}{18(x + 3)}$.

Step3: Integrate term - by - term

$\int\frac{x + 7}{2x^{3}-18x}dx=\int\left(-\frac{7}{18x}+\frac{5}{18(x - 3)}+\frac{1}{18(x + 3)}\right)dx=-\frac{7}{18}\int\frac{1}{x}dx+\frac{5}{18}\int\frac{1}{x - 3}dx+\frac{1}{18}\int\frac{1}{x + 3}dx$. Using the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have $=-\frac{7}{18}\ln|x|+\frac{5}{18}\ln|x - 3|+\frac{1}{18}\ln|x + 3|+C=\frac{1}{18}\left(-7\ln|x| + 5\ln|x - 3|+\ln|x + 3|\right)+C$.

Answer:

$\frac{x + 7}{2x^{3}-18x}=-\frac{7}{18x}+\frac{5}{18(x - 3)}+\frac{1}{18(x + 3)}$; $\int\frac{x + 7}{2x^{3}-18x}dx=\frac{1}{18}\left(-7\ln|x| + 5\ln|x - 3|+\ln|x + 3|\right)+C$