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

The key idea is to notice that the derivative of $\frac{e^x}{x}$ appears in the integrand.

Recall the quotient rule: $\frac{d}{dx}\left(\frac{e^x}{x}\right) = \frac{x e^x - e^x}{x^2} = e^x \left(\frac{1}{x} - \frac{1}{x^2}\right)$. …

The problem asks us to integrate $e^x \left( \frac{1}{x} - \frac{1}{x^2} \right)$. At first glance, this looks like a product of $e^x$ with a rational expression — not obviously a standard form. But there's a beautiful pattern hiding here.

The trick is to notice that $\frac{1}{x} - \frac{1}{x^2}$ is actually the derivative of $\frac{1}{x}$ plus $\frac{1}{x}$ itself. Let's check: if $f(x) = \frac{1}{x}$, then $f'(x) = -\frac{1}{x^2}$. So $f(x) + f'(x) = \frac{1}{x} - \frac{1}{x^2}$. That's exactly the bracket we have.

Why does this matter? Because there's a known result: $\int e^x \left[ f(x) + f'(x) \right] dx = e^x f(x) + C$. You can verify it by differentiating $e^x f(x)$ — the product rule gives $e^x f(x) + e^x f'(x) = e^x [f(x) + f'(x)]$, which is precisely the integrand. So integration undoes that differentiation.

This is a powerful shortcut. Instead of expanding or trying integration by parts, we just identify $f(x)$ and write the answer.

Let's apply it step by step.

1. Identify $f(x)$. We need a function $f(x)$ such that $f(x) + f'(x)$ matches the bracket $\frac{1}{x} - \frac{1}{x^2}$. Try $f(x) = \frac{1}{x}$. Then $f'(x) = -\frac{1}{x^2}$, and indeed $f(x) + f'(x) = \frac{1}{x} - \frac{1}{x^2}$. That's a perfect match.

2. Apply the formula. Since the integrand is $e^x \left[ f(x) + f'(x) \right]$ with $f(x) = \frac{1}{x}$, the integral is simply $e^x f(x) + C = e^x \cdot \frac{1}{x} + C$. …

Method: The $e^{x}\big(f(x)+f'(x)\big)$ shortcut

Whenever an integrand is $e^{x}$ times a bracket that is a function plus its own derivative, the answer is simply $e^{x}$ times that function:

Steps

Step 1: Isolate the $e^{x}$ factor and look at what multiplies it.

Step 2: Try to split the multiplier as $f(x)+f'(x)$.

Guess a candidate $f(x)$ (often the "nicer" of the two pieces, or the part left after a $\tfrac1x$-type term), then check that its derivative supplies the remaining piece. …

Common Mistakes

Mistake 1: Integrating $\dfrac1x$ and $\dfrac{1}{x^{2}}$ separately and mismatching.

Why it's wrong: the elegance is that $f=\dfrac1x$ has $f'=-\dfrac{1}{x^{2}}$, so $\dfrac1x-\dfrac{1}{x^{2}}=f+f'$. Correct approach: recognise the pattern instead of integrating piecewise.

Mistake 2: Picking $f=-\dfrac1x$ or $f=\dfrac{1}{x^{2}}$. …