Q.Integrate the following function: $\frac{x+3}{x^2 - 2x - 5}$

Split the numerator into (derivative of denominator) + constant.

$\dfrac{d}{dx}(x^2-2x-5) = 2x-2$, and $x+3 = \tfrac12(2x-2) + 4$.

First: $\tfrac12\log|x^2-2x-5|.$

Second: complete the square, $x^2-2x-5 = (x-1)^2-6$. With $\int \dfrac{du}{u^2-a^2} = \dfrac{1}{2a}\log\left|\dfrac{u-a}{u+a}\right|$ and $a=\sqrt6$: …

The strategy

For a linear numerator over a quadratic that doesn't factor over the rationals, force the numerator to contain the derivative of the denominator. Whatever is left over is a constant, handled by completing the square.

Match $x+3 = A(2x-2)+B$: from the $x$-term $2A=1\Rightarrow A=\tfrac12$; then $-2A+B = 3 \Rightarrow B=4$. So $x+3 = \tfrac12(2x-2)+4$.

Step 1 — the derivative part (a log)

since the numerator is exactly the derivative of the denominator.

Step 2 — the constant part (complete the square)

$x^2-2x-5 = (x-1)^2 - 6 = (x-1)^2 - (\sqrt6)^2.$ …

Method: $\dfrac{\text{linear}}{\text{quadratic}}$ with real roots — log part $+$ difference-of-squares log

For a linear numerator over a quadratic that has real roots (positive constant subtracted after completing the square), split into a derivative part and a constant part; the constant part is a difference of squares, giving a logarithm of a ratio — not an arctan.

Steps

Step 1: Split. With $D(x)=x^2+bx+c$ and $D'(x)=2x+b$, write $px+q=\lambda D'(x)+\mu$. …

Common Mistakes

Mistake 1: Using arctan for the constant piece.

Why it's wrong: $x^2-2x-5$ completes to $(x-1)^2-6$, a difference of squares, so the integral is $\tfrac{1}{2a}\log\big|\tfrac{u-a}{u+a}\big|$, not $\tfrac1a\tan^{-1}\tfrac{u}{a}$. Correct approach: check the sign — a negative constant after completing the square means the log-ratio form.

Mistake 2: Forgetting the $\tfrac{1}{2a}$ factor. …