Q.Integrate the following function: $\frac{x^2}{\sqrt{x^6+a^6}}$

Concept: U Substitution — we rewrite the integrand so that the chain rule in reverse simplifies the square root.

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

The integral becomes:

This is a standard form: $\int \frac{du}{\sqrt{u^2 + k^2}} = \sinh^{-1}\left(\frac{u}{k}\right) + C$ (or $\log|u + \sqrt{u^2 + k^2}| + C$). Here $k = a^3$.

Thus: …

We start with the integral

The denominator contains $x^6 + a^6$, and the numerator is $x^2$. Notice that $x^6 = (x^3)^2$, so the square root is $\sqrt{(x^3)^2 + a^6}$. This suggests that if we set $u = x^3$, then $du = 3x^2 \, dx$, and $x^2 \, dx$ appears almost exactly in the numerator — we just need a factor of $3$.

1. Substitution setup

   Let $u = x^3$. Then $du = 3x^2 \, dx$, so $x^2 \, dx = \frac{1}{3} du$.

   Also, $x^6 = (x^3)^2 = u^2$, so the denominator becomes $\sqrt{u^2 + a^6}$.

2. Rewrite the integral

   Substituting, we get

3. Recognize the standard form

   The integral $\int \frac{du}{\sqrt{u^2 + k^2}}$ is a standard result: it equals $\log\left| u + \sqrt{u^2 + k^2} \right| + C$.

   Here $k^2 = a^6$, so $k = a^3$ (taking the positive root, since $a$ is presumably real and $a^6$ is positive).

   $\int \frac{du}{\sqrt{u^2 + k^2}} = \log\left| u + \sqrt{u^2 + k^2} \right| + C$

4. Apply the formula

   With $k = a^3$, we have

5. Back-substitute Replace $u$ with $x^3$: …

Method: Substitute $u=x^3$ to reach $\displaystyle\int\frac{du}{\sqrt{u^2+a^2}}$

When $x^2\,dx$ appears with an $x^6$ under a radical, substitute $u=x^3$ (since $x^6=(x^3)^2$) to convert to a standard square-root form.

Steps

Step 1: Substitute $u=x^3$.

and $x^6=u^2$, $a^6=(a^3)^2$.

Step 2: Rewrite the integral.

Step 3: Apply the standard result and restore. …

Common Mistakes

Mistake 1: Choosing $u=x^6$ instead of $u=x^3$.

Why it's wrong: the numerator $x^2\,dx$ is $\tfrac13\,d(x^3)$, so $u=x^3$ is what the differential supports; $u=x^6$ leaves stray powers of $x$. Correct approach: set $u=x^3$, giving $x^2\,dx=\tfrac13\,du$ and $x^6=u^2$.

Mistake 2: Not recognising $a^6=(a^3)^2$ so the standard form applies. …