evaluate the following integral\n int\frac{x^{2}+10x - 1}{x^{3}-x}dx \nfind the partial fraction…

evaluate the following integral\n int\frac{x^{2}+10x - 1}{x^{3}-x}dx \nfind the partial fraction decomposition of the integrand\n int\frac{x^{2}+10x - 1}{x^{3}-x}dx=intleft(\frac{1}{x}+\frac{5}{x - 1}-\frac{5}{x + 1}\right)dx \nevaluate the indefinite integral\n int\frac{x^{2}+10x - 1}{x^{3}-x}dx=square \n(use parentheses to clearly denote the argument of each function.)

evaluate the following integral\n int\frac{x^{2}+10x - 1}{x^{3}-x}dx \nfind the partial fraction decomposition of the integrand\n int\frac{x^{2}+10x - 1}{x^{3}-x}dx=intleft(\frac{1}{x}+\frac{5}{x - 1}-\frac{5}{x + 1}\right)dx \nevaluate the indefinite integral\n int\frac{x^{2}+10x - 1}{x^{3}-x}dx=square \n(use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Recall integral rules

We know that $\int\frac{1}{x}dx=\ln|x|+C$, $\int\frac{1}{x - a}dx=\ln|x - a|+C$ and $\int\frac{1}{x + a}dx=\ln|x + a|+C$.

Step2: Integrate each term

Given $\int\left(\frac{1}{x}+\frac{5}{x - 1}-\frac{5}{x + 1}\right)dx=\int\frac{1}{x}dx+5\int\frac{1}{x - 1}dx-5\int\frac{1}{x + 1}dx$. Using the integral rules above, we have $\ln|x|+5\ln|x - 1|-5\ln|x + 1|+C$.

Step3: Simplify using logarithm properties

$5\ln|x - 1|-5\ln|x + 1| = 5(\ln|x - 1|-\ln|x + 1|)=5\ln\left|\frac{x - 1}{x + 1}\right|$. So the result is $\ln|x|+5\ln\left|\frac{x - 1}{x + 1}\right|+C$.

Answer:

$\ln|x|+5\ln\left|\frac{x - 1}{x + 1}\right|+C$