evaluate the following integral\n int\frac{88}{x^{2}-121}dx \n int\frac{88}{x^{2}-121}dx=square \n(use…

evaluate the following integral\n int\frac{88}{x^{2}-121}dx \n int\frac{88}{x^{2}-121}dx=square \n(use parentheses to clearly denote the argument of each function)

evaluate the following integral\n int\frac{88}{x^{2}-121}dx \n int\frac{88}{x^{2}-121}dx=square \n(use parentheses to clearly denote the argument of each function)

Answer

Explanation:

Step1: Factor the denominator

We know that $x^{2}-121=(x - 11)(x + 11)$ by the difference - of - squares formula $a^{2}-b^{2}=(a - b)(a + b)$ where $a=x$ and $b = 11$. Then we perform partial - fraction decomposition on $\frac{88}{x^{2}-121}=\frac{88}{(x - 11)(x + 11)}=\frac{A}{x - 11}+\frac{B}{x + 11}$. Cross - multiplying gives $88=A(x + 11)+B(x - 11)$. Let $x=11$, then $88=A(11 + 11)+B(0)$, so $A = 4$. Let $x=-11$, then $88=A(0)+B(-11 - 11)$, so $B=-4$. So $\frac{88}{x^{2}-121}=\frac{4}{x - 11}-\frac{4}{x + 11}$.

Step2: Integrate term - by - term

$\int\frac{88}{x^{2}-121}dx=\int(\frac{4}{x - 11}-\frac{4}{x + 11})dx$. Using the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have $\int\frac{4}{x - 11}dx-\int\frac{4}{x + 11}dx=4\ln|x - 11|-4\ln|x + 11|+C$. Using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we can rewrite it as $4\ln\left|\frac{x - 11}{x + 11}\right|+C$.

Answer:

$4\ln\left|\frac{x - 11}{x + 11}\right|+C$