Q.Integrate the following function: $\frac{5x-2}{1+2x+3x^2}$

The key idea is Integration by Completing the Square, combined with splitting the numerator to match the derivative of the denominator.

First, differentiate the denominator:

$\frac{d}{dx}(1+2x+3x^2) = 2 + 6x$.

Rewrite the numerator as a multiple of this derivative plus a constant:

$5x - 2 = \frac{5}{6}(6x+2) - \frac{11}{3}$.

Check: $\frac{5}{6}(6x+2) = 5x + \frac{5}{3}$, then subtract $\frac{11}{3}$ gives $5x - 2$.

Now the integral splits:

$\int \frac{5x-2}{1+2x+3x^2}dx = \frac{5}{6}\int \frac{6x+2}{1+2x+3x^2}dx - \frac{11}{3}\int \frac{dx}{1+2x+3x^2}$.

The first part is $\frac{5}{6}\log|1+2x+3x^2|$.

For the second, complete the square: $1+2x+3x^2 = 3\left(x^2 + \frac{2}{3}x + \frac{1}{3}\right) = 3\left[\left(x+\frac{1}{3}\right)^2 + \frac{2}{9}\right] = 3\left(x+\frac{1}{3}\right)^2 + \frac{2}{3}$. …

When you see a quadratic denominator like $1+2x+3x^2$, the first instinct is often to check if the numerator is a multiple of the derivative of the denominator. That would give a simple log. Here, the derivative of the denominator is $2+6x$, and our numerator is $5x-2$ — not a perfect match, but close. The trick is to complete the square in the denominator to turn it into something like $a^2 + (x+b)^2$, which then invites an arctan substitution for the leftover constant part.

Let’s walk through it cleanly.

1. Complete the square in the denominator

   We have $3x^2 + 2x + 1$. Factor out the $3$ from the quadratic terms:

Complete the square inside the bracket: $x^2 + \frac{2}{3}x = \left(x + \frac{1}{3}\right)^2 - \frac{1}{9}$.

So

Factor the constant to make it look like $a^2 + u^2$:

So the denominator becomes $3\left[(x + \frac{1}{3})^2 + \left(\frac{\sqrt{2}}{3}\right)^2\right]$.

2. Rewrite the integral

3. Split the numerator to match the derivative of the denominator

   The derivative of $(x+\frac{1}{3})^2 + \frac{2}{9}$ is $2(x+\frac{1}{3}) = 2x + \frac{2}{3}$. We want to express $5x-2$ as $A(2x+\frac{2}{3}) + B$.

   Write:

Compare coefficients of $x$: $5 = 2A \implies A = \frac{5}{2}$.

Compare constant terms: $-2 = A\cdot\frac{2}{3} + B = \frac{5}{2}\cdot\frac{2}{3} + B = \frac{5}{3} + B \implies B = -2 - \frac{5}{3} = -\frac{11}{3}$.

So

4. Substitute back into the integral

Split into two integrals:

Simplify the constants:

5. First integral: log form

   Notice that the numerator $2x+\frac{2}{3}$ is exactly the derivative of the denominator $(x+\frac{1}{3})^2 + \frac{2}{9}$. So

But $(x+\frac{1}{3})^2 + \frac{2}{9} = \frac{1}{3}(1+2x+3x^2)$, so the log is $\log\left|\frac{1}{3}(1+2x+3x^2)\right| = \log|1+2x+3x^2| - \log 3$. The constant $-\log 3$ gets absorbed into $C$, so we can simply write $\log|1+2x+3x^2|$. …

Method: $\dfrac{\text{linear}}{\text{quadratic}}$ — split into a log part and an arctan part

For a linear numerator over a quadratic with no real roots, split the numerator into (a multiple of the derivative of the denominator) + (a constant); the first gives a logarithm, the second an arctangent.

Steps

Step 1: Split the numerator. With denominator $D(x)=ax^2+bx+c$ and $D'(x)=2ax+b$,

Step 2: First piece — log of the denominator.

$\lambda\int \frac{D'(x)}{D(x)}\,dx = \lambda\log|D(x)| .$ …

Common Mistakes

Mistake 1: Forgetting the leading coefficient when completing the square.

Why it's wrong: for $3x^2+2x+1$ you must first factor out the $3$; ignoring it scales the arctan term wrongly. Correct approach: write $D(x)=a\big[(x+h)^2+k^2\big]$ and keep the $\tfrac1a$ outside.

Mistake 2: Using a $\log$-of-difference form for the constant piece. …