Question

calculate the integral: $int\frac{1}{(x + 5)(x + 8)}dx$

Explanation:

Step1: Decompose into partial - fractions

Let $\frac{1}{(x + 5)(x + 8)}=\frac{A}{x + 5}+\frac{B}{x + 8}$. Then $1=A(x + 8)+B(x + 5)$. Let $x=-5$, we get $1 = 3A$, so $A=\frac{1}{3}$. Let $x=-8$, we get $1=-3B$, so $B =-\frac{1}{3}$. So $\frac{1}{(x + 5)(x + 8)}=\frac{1}{3(x + 5)}-\frac{1}{3(x + 8)}$.

Step2: Integrate term - by - term

$\int\frac{1}{(x + 5)(x + 8)}dx=\int(\frac{1}{3(x + 5)}-\frac{1}{3(x + 8)})dx=\frac{1}{3}\int\frac{1}{x + 5}dx-\frac{1}{3}\int\frac{1}{x + 8}dx$.
Using the formula $\int\frac{1}{u}du=\ln|u|+C$, we have $\frac{1}{3}\ln|x + 5|-\frac{1}{3}\ln|x + 8|+C$.

Step3: Simplify the result

Using the property $\ln a-\ln b=\ln\frac{a}{b}$, we get $\frac{1}{3}\ln|\frac{x + 5}{x + 8}|+C$.

Answer:

$\frac{1}{3}\ln|\frac{x + 5}{x + 8}|+C$