Question

calculate the integral

1/(x^2+14x+48)

Answer 1

factor, then use partial fractions

Answer 2

\[\int\frac1{x^2+14x+48}dx=\int\frac1{(x+6)(x+8)}dx=\int\frac A{x+6}+\frac B{x+8}dx\]\[A(x+8)+B(x+6)=1\]\[x=-6\to A=\frac12\]\[x=-8\to B=-\frac12\]\[\int\frac12\frac1{x+6}-\frac12\frac1{x+8}dx=\frac12\ln|x+6|-\frac12\ln|x+8|+C\]