Q.Integrate the following function: $\frac{(x-3)e^x}{(x-1)^3}$

The key idea is to recognise the integrand as the derivative of a quotient, which lets us integrate directly using the reverse of the quotient rule (sometimes called the Power Rule for Integration in the form $\int \frac{u'v - uv'}{v^2} = \frac{u}{v} + C$).

We rewrite the given function as:

Now split into two fractions:

$\frac{e^x}{(x-1)^2} - \frac{2e^x}{(x-1)^3}$ …

Why This Approach Works

When you see an integrand of the form $e^x$ times a rational function, your first instinct should be to check if it matches the derivative of $e^x f(x)$. The product rule gives:

So if we can express our integrand as $e^x$ times something that looks like $f(x) + f'(x)$, the integral is simply $e^x f(x) + C$. This is the reverse of the product rule — a technique often called "integration by recognition" or "the $e^x$ trick."

Our integrand is $\frac{(x-3)e^x}{(x-1)^3}$. The $e^x$ factor is already there, so we need to find a function $f(x)$ such that:

The denominator $(x-1)^3$ suggests $f(x)$ might be of the form $\frac{A}{(x-1)^2}$ or $\frac{B}{(x-1)}$ — let's try the simplest guess.

So the guess works perfectly. No trial and error needed — the pattern jumps out once you check.

Step-by-Step Solution

1. Recognize the pattern.

   We want to find $f(x)$ such that $\frac{d}{dx}\left(e^x f(x)\right) = e^x \cdot \frac{x-3}{(x-1)^3}$.

   This means $f(x) + f'(x) = \frac{x-3}{(x-1)^3}$.

2. Guess $f(x)$ from the denominator.

   Since the denominator is $(x-1)^3$, try $f(x) = \frac{1}{(x-1)^2}$.

   Compute $f'(x)$: …

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: Not rewriting $\dfrac{x-3}{(x-1)^{3}}$ as $f+f'$.

Why it's wrong: it equals $\dfrac{1}{(x-1)^{2}}-\dfrac{2}{(x-1)^{3}}$, i.e. $f+f'$ with $f=\dfrac{1}{(x-1)^{2}}$. Correct approach: split the numerator ($x-3=(x-1)-2$) before integrating.

Mistake 2: Differentiating $\dfrac{1}{(x-1)^{2}}$ incorrectly. …