Q.Integrate the function $\frac{3x^2}{x^6+1}$

The key idea is U Substitution: the numerator $3x^2$ is almost the derivative of $x^3$, which appears inside the denominator.

Let $u = x^3$. Then $du = 3x^2 \, dx$, so the integral becomes

This is a standard form: $\int \frac{du}{u^2+1} = \tan^{-1} u + C$.

Substitute back $u = x^3$ to get the final antiderivative.

The key to this problem is recognizing that the numerator is almost the derivative of the denominator's "inner" part. The denominator is $x^6+1$, which is $(x^3)^2 + 1$. If we set $u = x^3$, then $du = 3x^2 dx$ — and that's exactly the numerator! This is a textbook case for U Substitution: we look for a function and its derivative hiding in the integrand.

Let's walk through it step by step.

1. Identify the substitution. The denominator $x^6+1$ can be written as $(x^3)^2 + 1$. This suggests letting $u = x^3$. Why? Because the derivative of $x^3$ is $3x^2$, which appears in the numerator. So set:

2. Compute the differential. Differentiate both sides:

Notice that $3x^2 dx$ is exactly the numerator of the integrand. This is perfect — the substitution will replace the entire numerator and $dx$ in one go.

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

Replace $3x^2 dx$ with $du$, and $x^6$ with $(x^3)^2 = u^2$:

4. Integrate using a standard formula. The integral $\int \frac{du}{u^2 + a^2}$ is $\frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$. Here $a = 1$, so:

5. Substitute back to $x$. Since $u = x^3$, we replace $u$:

Method: Substitution Recognising "Derivative-in-the-Numerator"

Use this when the numerator is (a constant times) the derivative of an inner expression that appears in the denominator — a hallmark of reverse chain rule leading to a standard form.

Steps

Step 1: Rewrite the denominator to reveal the inner function.

Look for a perfect power. Here $x^6+1=(x^3)^2+1$, which suggests the inner function $u=x^3$.

Step 2: Check the numerator against $du$.

With $u=x^3$, $du=3x^2\,dx$ — exactly the numerator $3x^2\,dx$. When the numerator matches $du$, the substitution collapses the integral cleanly:

Step 3: Apply the standard form and back-substitute.

Recognise $\int\dfrac{du}{u^2+1}=\tan^{-1}u+C$, then restore $u=x^3$:

Common Mistakes

Mistake 1: Not recognising $x^6=(x^3)^2$.

Why it's wrong: missing this hides the $u=x^3$ substitution and makes the integral look intractable. Correct approach: rewrite even-power denominators as squares to spot the arctan form.

Mistake 2: Confusing $\dfrac{1}{u^2+1}$ with a logarithm.

Why it's wrong: $\int\frac{du}{u^2+1}=\tan^{-1}u$, whereas the log form needs $\frac{1}{u^2-1}$ or $\frac{f'}{f}$. Correct approach: memorise $\int\frac{du}{u^2+1}=\tan^{-1}u+C$.

Mistake 3: Introducing a stray constant factor.

Why it's wrong: since $du=3x^2\,dx$ matches the numerator exactly, no extra $\frac13$ is needed. Correct approach: only insert a compensating constant when the numerator is a multiple of $du$, not an exact match.