Q.Evaluate: $\int \frac{3x+4}{x^2+6x+5}\,dx$

When the numerator of a quadratic-denominator (or quadratic-under-a-root) integral is linear rather than constant, it is first rewritten as $px+q=A\cdot\frac{d}{dx}(ax^2+bx+c)+B$, with $A,B$ found by comparing coefficients (or, for square-root denominators, $\frac{d}{dx}\sqrt{ax^2+bx+c}$ analogues). This splits the integral into an $A$-piece, which is a pure "log of the denominator" (or, under a root, "square root of the denominator") found by the substitution $t=ax^2+bx+c$, plus a $B$-piece of the standard quadratic-denominator type solved by completing the square. The identical idea, applied after first rationalising a surd ratio like $\sqrt{\frac{x-a}{x-b}}$ by multiplying inside the root by the same factor top and bottom, also handles integrals whose linear-over-square-root shape is disguised inside a fraction of two linear expressions under a single square root.