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

express the integrand as a sum of partial fractions and evaluate the integrals.\n int\frac{x + 4}{2x^{3}-8x}dx\nrewrite the integrand as the sum of partial fractions.\n \frac{x + 4}{2x^{3}-8x}=\nevaluate the integrals.\n int\frac{x + 4}{2x^{3}-8x}dx=\n(use parentheses to clearly denote the argument of each function.)

express the integrand as a sum of partial fractions and evaluate the integrals.\n int\frac{x + 4}{2x^{3}-8x}dx\nrewrite the integrand as the sum of partial fractions.\n \frac{x + 4}{2x^{3}-8x}=\nevaluate the integrals.\n int\frac{x + 4}{2x^{3}-8x}dx=\n(use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Factor the denominator

First, factor out the common factor from the denominator $2x^{3}-8x = 2x(x^{2}-4)=2x(x - 2)(x + 2)$.

Step2: Set up partial - fraction decomposition

Let $\frac{x + 4}{2x^{3}-8x}=\frac{x + 4}{2x(x - 2)(x + 2)}=\frac{A}{x}+\frac{B}{x - 2}+\frac{C}{x + 2}$. Then $x + 4=A(x - 2)(x + 2)+B(2x)(x + 2)+C(2x)(x - 2)$.

Step3: Find the values of A, B, and C

If $x = 0$: $0+4=A(0 - 2)(0 + 2)+B\times0+C\times0$, so $4=-4A$, and $A=-1$. If $x = 2$: $2 + 4=A\times0+B(2\times2)(2 + 2)+C\times0$, so $6 = 16B$, and $B=\frac{3}{8}$. If $x=-2$: $-2 + 4=A\times0+B\times0+C(2\times(-2))(-2 - 2)$, so $2 = 16C$, and $C=\frac{1}{8}$. So $\frac{x + 4}{2x^{3}-8x}=-\frac{1}{2x}+\frac{3}{16(x - 2)}+\frac{1}{16(x + 2)}$.

Step4: Evaluate the integral

$\int\frac{x + 4}{2x^{3}-8x}dx=\int\left(-\frac{1}{2x}+\frac{3}{16(x - 2)}+\frac{1}{16(x + 2)}\right)dx$ $=-\frac{1}{2}\int\frac{1}{x}dx+\frac{3}{16}\int\frac{1}{x - 2}dx+\frac{1}{16}\int\frac{1}{x + 2}dx$ $=-\frac{1}{2}\ln|x|+\frac{3}{16}\ln|x - 2|+\frac{1}{16}\ln|x + 2|+C$.

Answer:

$\frac{x + 4}{2x^{3}-8x}=-\frac{1}{2x}+\frac{3}{16(x - 2)}+\frac{1}{16(x + 2)}$; $\int\frac{x + 4}{2x^{3}-8x}dx=-\frac{1}{2}\ln|x|+\frac{3}{16}\ln|x - 2|+\frac{1}{16}\ln|x + 2|+C$