Q.Find the following integrals:

(i) Complete the square: $x^2-6x+13=(x-3)^2+2^2$. Then

(ii) Factor $3x^2+13x-10=(3x-2)(x+5)$ and use partial fractions: $\dfrac{1}{(3x-2)(x+5)}=\dfrac{3/17}{3x-2}-\dfrac{1/17}{x+5}$. Integrating,

(iii) Inside the root, $5x^2-2x=5\left[(x-\tfrac15)^2-(\tfrac15)^2\right]$, so …

(i) $\int\dfrac{dx}{x^2-6x+13}$

The denominator has no real roots, so complete the square:

With $u=x-3$ and $\int\dfrac{du}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac{u}{a}$ ($a=2$):

(ii) $\int\dfrac{dx}{3x^2+13x-10}$

This quadratic factors, so partial fractions are cleanest. $3x^2+13x-10=(3x-2)(x+5)$. Write

Put $x=\tfrac23$: $1=A\cdot\tfrac{17}{3}\Rightarrow A=\tfrac{3}{17}$. Put $x=-5$: $1=B(-17)\Rightarrow B=-\tfrac{1}{17}$. Then

(iii) $\int\dfrac{dx}{\sqrt{5x^2-2x}}$

Factor out $5$ and complete the square inside the root:

$5x^2-2x=5\left(x^2-\frac25x\right)=5\left[\left(x-\frac15\right)^2-\frac{1}{25}\right],$ …

Method: Completing the Square for Quadratic Denominators

Use this for $\dfrac{1}{ax^2+bx+c}$ and $\dfrac{1}{\sqrt{ax^2+bx+c}}$ integrals: reshape the quadratic into $(x-h)^2\pm k^2$ so a standard form applies.

Steps

Step 1: Complete the square (and factor out any leading coefficient first).

Write $ax^2+bx+c=a\big[(x+\tfrac{b}{2a})^2+\text{const}\big]$. For example $x^2-6x+13=(x-3)^2+4$.

Step 2: Decide the standard form from the sign of the constant.

• $(x-h)^2+k^2$ in a denominator → $\dfrac1k\tan^{-1}\dfrac{x-h}{k}$. …

Common Mistakes

Mistake 1: Completing the square without factoring out the leading coefficient.

Why it's wrong: for $3x^2+13x-10$ you must handle the coefficient $3$ (here it factors, so use partial fractions); ignoring it corrupts the constant. Correct approach: factor out or divide by the leading coefficient first.

Mistake 2: Using arctan when the quadratic factors. …