Q.Evaluate:
$\int_1^2 \frac{1}{x^2 + 6x + 5} \, dx$

$x^2+6x+5=(x+1)(x+5)$, and $\dfrac{1}{(x+1)(x+5)}=\dfrac14\!\left(\dfrac{1}{x+1}-\dfrac{1}{x+5}\right)$, so the integral $=\dfrac14\Big[\ln\big|\tfrac{x+1}{x+5}\big|\Big]_1^2=\dfrac14\ln\dfrac97$.

Step 1 — factorise the denominator:

$x^2+6x+5=(x+1)(x+5).$

Step 2 — partial fractions. Write $\dfrac{1}{(x+1)(x+5)}=\dfrac{A}{x+1}+\dfrac{B}{x+5}$, so $1=A(x+5)+B(x+1)$.

Put $x=-1$: $1=4A\Rightarrow A=\tfrac14$. Put $x=-5$: $1=-4B\Rightarrow B=-\tfrac14$.

$\frac{1}{(x+1)(x+5)}=\frac14\left(\frac{1}{x+1}-\frac{1}{x+5}\right).$

Step 3 — integrate. …