Q.Integrate the following function: $\frac{1}{9x^2 + 6x + 5}$

Concept: U Substitution — we complete the square in the denominator to match the standard form $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\frac{x}{a} + C$.

Step 1: Complete the square.

$9x^2 + 6x + 5 = 9\left(x^2 + \frac{2}{3}x\right) + 5 = 9\left[\left(x + \frac{1}{3}\right)^2 - \frac{1}{9}\right] + 5 = 9\left(x + \frac{1}{3}\right)^2 + 4$.

Step 2: Rewrite the integral.

Step 3: Substitute $u = x + \frac{1}{3}$, so $du = dx$.

$= \int \frac{du}{9u^2 + 4} = \frac{1}{4} \int \frac{du}{\left(\frac{3u}{2}\right)^2 + 1}$ …

We are integrating $\int \frac{1}{9x^2 + 6x + 5} \, dx$. The denominator is a quadratic that does not factor nicely over the reals (its discriminant is $6^2 - 4\cdot9\cdot5 = 36 - 180 = -144 < 0$). This immediately suggests the integral will involve an inverse tangent.

The standard form for such integrals is:

So our job is to force the denominator into the shape $u^2 + a^2$ by completing the square.

Why completing the square works here: A quadratic $Ax^2 + Bx + C$ can always be written as $A(x + \frac{B}{2A})^2 + (C - \frac{B^2}{4A})$. This isolates the variable into a single squared term, leaving a constant. Once we have $(something)^2 + constant$, a simple substitution $u = something$ and a factor adjustment gives us the arctangent form.

Let's do it step by step.

1. Complete the square in the denominator. Factor out the coefficient of $x^2$ from the first two terms:

Inside the parentheses, take half of $\frac{2}{3}$ (which is $\frac{1}{3}$) and square it to get $\frac{1}{9}$. Add and subtract this inside:

Distribute the 9:

So the denominator becomes $9(x + \frac{1}{3})^2 + 4$.

2. Factor to match the $u^2 + a^2$ pattern.

   We have $9(x + \frac{1}{3})^2 + 4$. Notice that $9(x + \frac{1}{3})^2 = [3(x + \frac{1}{3})]^2 = (3x + 1)^2$.

   So the denominator is $(3x+1)^2 + 4$.

   This is exactly $u^2 + a^2$ with $u = 3x+1$ and $a = 2$ (since $2^2 = 4$).

3. Perform the substitution.

   Let $u = 3x + 1$. Then $du = 3 \, dx$, so $dx = \frac{du}{3}$.

   The integral becomes:

4. Apply the arctangent formula. …

Method: Complete the square, then use $\displaystyle\int\frac{du}{u^2+a^2}$ (arctangent)

For $\dfrac{1}{\text{quadratic}}$ where the quadratic has no real roots, complete the square to reach $(\text{linear})^2+a^2$ in the denominator, then apply the arctangent standard form.

Steps

Step 1: Complete the square in the denominator.

(check: $(3x+1)^2=9x^2+6x+1$, plus $4$ gives $9x^2+6x+5$).

Step 2: Substitute the linear block.

$u=3x+1$, $du=3\,dx\Rightarrow dx=\tfrac13\,du$, so the integral is $\dfrac13\displaystyle\int\frac{du}{u^2+4}$.

Step 3: Apply the standard result. …

Common Mistakes

Mistake 1: Forgetting the $\dfrac13$ from $dx=\tfrac13\,du$ after substituting $u=3x+1$.

Why it's wrong: because $u=3x+1$ has slope $3$, $dx=\tfrac13\,du$; dropping it makes the answer three times too large. Correct approach: carry the $\tfrac13$ alongside the $\tfrac1a=\tfrac12$, giving the $\tfrac16$ coefficient.

Mistake 2: Applying a $\log$/$\sinh^{-1}$ form instead of $\tan^{-1}$. …