Q.Integrate the following function: $\int x^2 \left(1 - \frac{1}{x^2}\right) dx$

The key idea is to expand the integrand first, turning the product into simpler power terms that can be integrated term-by-term.

Step 1: Expand the bracket:

Step 2: Integrate each term separately:

$\int (x^2 - 1) \, dx = \int x^2 \, dx - \int 1 \, dx$ …

Why “Integration by Expansion” works here

When you see a product like $x^2$ times a bracket, your first instinct might be to look for a substitution. But look closely: the bracket itself is a simple polynomial in $1/x^2$. Multiplying through turns the whole thing into a sum of power functions — and integrating powers is the most straightforward operation in calculus. No chain rule, no substitution, no integration by parts. Just expand, then apply $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ for each term.

This is a classic “simplify before you differentiate (or integrate)” move. Many students rush to integrate a product without checking whether it can be expanded first. Here, expansion reduces the problem to two trivial integrals.

1. Expand the integrand Multiply $x^2$ into the bracket:

The $x^2$ cancels with the $1/x^2$, leaving a constant $-1$. So the integral becomes

2. Integrate term by term Use the power rule for $x^2$:

For the constant $-1$, recall that $\int -1 \, dx = -x$ (since $\int k\, dx = kx$ for any constant $k$).

3. Add the constant of integration Every indefinite integral must include $+C$: …

Method: Simplify a product or fraction before integrating

Always check whether an integrand simplifies algebraically — a product like $x^2\left(1-\tfrac{1}{x^2}\right)$ collapses to a polynomial.

Steps

Step 1: Multiply out / simplify.

Distribute: $x^2\cdot 1-x^2\cdot\dfrac{1}{x^2}=x^2-1$; the awkward term cancels.

Step 2: Integrate term by term. …

Common Mistakes

Mistake 1: Trying substitution or parts on $x^2\left(1-\frac{1}{x^2}\right)$.

Why it's wrong: a substitution like $u=1-\frac{1}{x^2}$ is needlessly messy. Correct approach: distribute first — it simplifies to $x^2-1$, then integrate to $\frac{x^3}{3}-x+C$. …