find the integral of $int\frac{dx}{x^{2}-4}$. \n$lnsqrt4{\frac{x + 2}{x - 2}}+c$\n$\frac{1}{4}ln(x +…

Explanation:

Step1: Decompose the fraction

We use partial - fraction decomposition. Since $x^{2}-4=(x + 2)(x - 2)$, we assume $\frac{1}{x^{2}-4}=\frac{A}{x + 2}+\frac{B}{x - 2}$. Then $1=A(x - 2)+B(x + 2)$. Let $x = 2$, we get $1=B(2 + 2)$, so $B=\frac{1}{4}$. Let $x=-2$, we get $1=A(-2 - 2)$, so $A=-\frac{1}{4}$. So $\frac{1}{x^{2}-4}=\frac{1}{4}\left(\frac{1}{x - 2}-\frac{1}{x + 2}\right)$.

Step2: Integrate term - by - term

$\int\frac{1}{x^{2}-4}dx=\frac{1}{4}\int\left(\frac{1}{x - 2}-\frac{1}{x + 2}\right)dx$. According to the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have $\frac{1}{4}\left(\int\frac{1}{x - 2}dx-\int\frac{1}{x + 2}dx\right)=\frac{1}{4}(\ln|x - 2|-\ln|x + 2|)+C$.

Step3: Use logarithm property

Using the property $\ln a-\ln b=\ln\frac{a}{b}$, we get $\frac{1}{4}\ln\left|\frac{x - 2}{x + 2}\right|+C=\ln\sqrt[4]{\left|\frac{x - 2}{x + 2}\right|}+C$.

Answer:

$\ln\sqrt[4]{\left|\frac{x - 2}{x + 2}\right|}+C$