Q.Evaluate the following.
$\int \frac{1}{7 + 6x - x^2} \, dx$

Complete the square in the denominator to get $16 - (x-3)^2$, then apply the standard integral $\displaystyle\int \frac{du}{a^2 - u^2} = \frac{1}{2a}\log\left|\frac{a+u}{a-u}\right| + C$ with $a=4,\ u=x-3$.

Rewrite the denominator:

$7 + 6x - x^2 = -\left(x^2 - 6x - 7\right) = -\left[(x-3)^2 - 16\right] = 16 - (x-3)^2.$

So the integral becomes

$\int \frac{dx}{16 - (x-3)^2}.$

Put $u = x-3$ (so $du = dx$) and $a = 4$. Using the standard form

$\int \frac{du}{a^2 - u^2} = \frac{1}{2a}\log\left|\frac{a+u}{a-u}\right| + C,$

we get …