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

We first expand the square, then integrate term by term using the power rule.

Now integrate each term:

$\int x \, dx = \frac{x^2}{2}, \quad \int (-2) \, dx = -2x, \quad \int \frac{1}{x} \, dx = \log|x|$ …

We start with the integral:

The instinct might be to try a substitution, but that would be messy. Instead, notice that the integrand is a square of a binomial. Expanding it will turn it into a sum of simple power functions — and those are the easiest things to integrate.

Recall the Power Rule for integration: for any real number $n \neq -1$,

And when $n = -1$, we have the special case $\int x^{-1} dx = \log|x| + C$.

So our plan is: expand, rewrite every term as $x^n$, then integrate each term separately.

1. Expand the square. Using $(a - b)^2 = a^2 - 2ab + b^2$, with $a = \sqrt{x}$ and $b = \frac{1}{\sqrt{x}}$:

2. Simplify each term.

   • $(\sqrt{x})^2 = x$
   • $(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) = 1$, so the middle term is $-2(1) = -2$
   • $\left(\frac{1}{\sqrt{x}}\right)^2 = \frac{1}{x}$

   So the integrand becomes:

3. Rewrite in power form.

   • $x$ is $x^1$
   • $-2$ is $-2x^0$ (since $x^0 = 1$)
   • $\frac{1}{x}$ is $x^{-1}$

   So we are integrating:

4. Integrate term by term.
   • For $x^1$: $n = 1$, so $\int x^1 dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}$ …

Method: Expand a squared-radical binomial

Integrands like $\left(\sqrt x-\tfrac{1}{\sqrt x}\right)^2$ become elementary once squared out.

Steps

Step 1: Expand with $(a-b)^2=a^2-2ab+b^2$.

$\left(\sqrt x-\tfrac{1}{\sqrt x}\right)^2=x-2+\dfrac1x$; the cross term is the constant $-2$.

Step 2: Rewrite in power form.

$x=x^1$, $-2=-2x^0$, $\dfrac1x=x^{-1}$. …

Common Mistakes

Mistake 1: Mis-squaring the cross term.

Why it's wrong: $-2\cdot\sqrt x\cdot\frac{1}{\sqrt x}=-2$, a constant — not $-\frac{2}{x}$ or $-2x$. Correct approach: expand to $x-2+\frac1x$.

Mistake 2: Dropping the absolute value in $\int\frac1x\,dx$. …