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

The key idea is U Substitution — the denominator’s inner function $2+3x^3$ has a derivative that appears in the numerator.

Let $u = 2 + 3x^3$. Then $du = 9x^2 \, dx$, so $x^2 \, dx = \frac{du}{9}$.

The integral becomes:

Integrate: …

The key insight here is that the numerator $x^2$ looks like it could be the derivative of something in the denominator — specifically, the derivative of $2+3x^3$ is $9x^2$. That’s almost a perfect match, just off by a constant factor. This is the classic signal for U Substitution: when you see a function and its derivative (up to a constant) multiplied together, substitution lets you collapse the whole expression into a simple power rule.

Let’s walk through it.

1. Choose the substitution.

   Let $u = 2 + 3x^3$. Why? Because the denominator is $(2+3x^3)^3$, and the derivative of $2+3x^3$ is $9x^2$, which is a constant multiple of the numerator $x^2$. This means $du$ will absorb the $x^2$ term cleanly.

2. Compute $du$ in terms of $dx$.

   Differentiate:

We have $x^2 \, dx$ in the integral, so solve for it:

3. Rewrite the integral entirely in $u$. The denominator $(2+3x^3)^3$ becomes $u^3$. So:

4. Integrate using the power rule. For $n \neq -1$, $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$. Here $n = -3$, so: $\frac{1}{9} \cdot \frac{u^{-2}}{-2} + C = -\frac{1}{18} u^{-2} + C.$ …

Method: Substitute the base of a power in the denominator

Use this when a power of $x$ out front matches the derivative of the bracket in the denominator, e.g. $\dfrac{x^2}{(2+3x^3)^3}$ where $\dfrac{d}{dx}(2+3x^3)=9x^2$.

Steps

Step 1: Substitute the bracket.

Let $u=2+3x^3$, so $du=9x^2\,dx$, i.e. $x^2\,dx=\dfrac{du}{9}$.

Step 2: Rewrite as a negative power of $u$.

$\int\frac{x^2}{(2+3x^3)^3}\,dx=\frac19\int u^{-3}\,du.$ …

Common Mistakes

Mistake 1: Not rewriting $\dfrac{1}{(2+3x^3)^3}$ as $(2+3x^3)^{-3}$.

Why it's wrong: the power rule needs an exponent form; left as a fraction the integral looks intractable. Correct approach: use a negative exponent, then substitute.

Mistake 2: Forgetting the $\dfrac19$ from $du=9x^2\,dx$.

Why it's wrong: with $u=2+3x^3$, $x^2\,dx=\dfrac{du}{9}$; dropping it makes the answer nine times too big. Correct approach: solve for $x^2\,dx$ before integrating. …