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

express the integrand as a sum of partial fractions and evaluate the integrals. (int\frac{x + 4}{2x^{3}-8x}dx) rewrite the integrand as the sum of partial fractions. (\frac{x + 4}{2x^{3}-8x}=-\frac{1}{2x}+\frac{3}{8(x - 2)}+\frac{1}{8(x + 2)}) evaluate the integrals. (int\frac{x + 4}{2x^{3}-8x}dx=square) (use parentheses to clearly denote the argument of each function.)

express the integrand as a sum of partial fractions and evaluate the integrals. (int\frac{x + 4}{2x^{3}-8x}dx) rewrite the integrand as the sum of partial fractions. (\frac{x + 4}{2x^{3}-8x}=-\frac{1}{2x}+\frac{3}{8(x - 2)}+\frac{1}{8(x + 2)}) evaluate the integrals. (int\frac{x + 4}{2x^{3}-8x}dx=square) (use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Split the integral

We know that $\int\frac{x + 4}{2x^{3}-8x}dx=\int\left(-\frac{1}{2x}+\frac{3}{8(x - 2)}+\frac{1}{8(x + 2)}\right)dx$. By the linear - ity of integration $\int(f(x)+g(x)+h(x))dx=\int f(x)dx+\int g(x)dx+\int h(x)dx$. So it becomes $-\frac{1}{2}\int\frac{1}{x}dx+\frac{3}{8}\int\frac{1}{x - 2}dx+\frac{1}{8}\int\frac{1}{x + 2}dx$.

Step2: Apply the integral formula

The integral formula for $\int\frac{1}{u}du=\ln|u|+C$. For $-\frac{1}{2}\int\frac{1}{x}dx=-\frac{1}{2}\ln|x|+C_1$. For $\frac{3}{8}\int\frac{1}{x - 2}dx=\frac{3}{8}\ln|x - 2|+C_2$. For $\frac{1}{8}\int\frac{1}{x + 2}dx=\frac{1}{8}\ln|x + 2|+C_3$.

Step3: Combine the results

Combining the above results, we get $-\frac{1}{2}\ln|x|+\frac{3}{8}\ln|x - 2|+\frac{1}{8}\ln|x + 2|+C$, where $C = C_1+C_2+C_3$.

Answer:

$-\frac{1}{2}\ln|x|+\frac{3}{8}\ln|x - 2|+\frac{1}{8}\ln|x + 2|+C$