Q.Integrate the following function: $\frac{x}{9-4x^2}$

The key idea is U Substitution — we look for a function and its derivative inside the integrand.

Notice that the derivative of $9 - 4x^2$ is $-8x$, which is a constant multiple of the numerator $x$. Let $u = 9 - 4x^2$. Then $du = -8x \, dx$, so $x \, dx = -\frac{1}{8} du$.

Substitute into the integral: …

The key insight here is pattern recognition. When you see a rational function where the numerator looks like the derivative of the denominator (up to a constant factor), your brain should immediately light up: u-substitution.

Look at the denominator: $9 - 4x^2$. Its derivative is $-8x$. The numerator is just $x$. That’s almost a perfect match — we’re only off by a factor of $-8$. This means we can set $u$ equal to the denominator, and the integral will collapse into a simple logarithmic form.

Let’s walk through it.

1. Choose the substitution.

   Let $u = 9 - 4x^2$.

   Why? Because the derivative $du = -8x \, dx$ contains an $x \, dx$ term, which is exactly what’s in the numerator (just missing a constant).

2. Solve for $x \, dx$.

   From $du = -8x \, dx$, we get

This is the cleanest way to replace the messy $x$ and $dx$ together.

3. Rewrite the integral entirely in terms of $u$. The original integral is

Substituting $u$ for the denominator and $-\frac{1}{8} du$ for $x \, dx$ gives:

4. Integrate with respect to $u$. The integral of $\frac{1}{u}$ is $\log|u| + C$. So:

5. Substitute back to $x$. …

Method: Substitute the quadratic denominator to get a logarithm

Use this when a single $x$ sits over a quadratic like $9-4x^2$, whose derivative $-8x$ matches the numerator up to a constant.

Steps

Step 1: Check the derivative of the denominator.

$\dfrac{d}{dx}(9-4x^2)=-8x$; the numerator $x$ is $-\tfrac18$ of it.

Step 2: Substitute $u=$ denominator. …

Common Mistakes

Mistake 1: Forgetting the $-\dfrac18$ from $du=-8x\,dx$.

Why it's wrong: with $u=9-4x^2$, $x\,dx=-\dfrac{du}{8}$; dropping the factor and sign gives a wrong coefficient. Correct approach: solve $du=-8x\,dx$ for $x\,dx$ before integrating.

Mistake 2: Losing the minus sign. …