Q.$\int e^{x}\left(\dfrac{1-x}{1+x^2}\right)^2 dx$ is equal to
(A) $\dfrac{e^{x}}{1+x^2} + C$
(B) $-\dfrac{e^{x}}{1+x^2} + C$
(C) $\dfrac{e^{x}}{(1+x^2)^2} + C$
(D) $-\dfrac{e^{x}}{(1+x^2)^2} + C$

Concept: Integration of Exponential — the integrand is of the form $e^x [f(x) + f'(x)]$, whose integral is $e^x f(x) + C$.

First, expand the square:

Now split the numerator:

Notice that

$\frac{d}{dx}\left(\frac{1}{1+x^2}\right) = -\frac{2x}{(1+x^2)^2}.$ …

The key to this problem is recognising a pattern: integrals of the form $\int e^x [f(x) + f'(x)]\,dx$ collapse beautifully to $e^x f(x) + C$. This is because the derivative of $e^x f(x)$ is $e^x f(x) + e^x f'(x)$, exactly the integrand. So when you see $e^x$ multiplied by something that looks like a function plus its derivative, you’re almost done.

Here, the integrand is $e^x \left(\frac{1-x}{1+x^2}\right)^2$. That square looks messy, but it might hide a neat structure. Let’s expand and see if we can spot $f(x) + f'(x)$.

1. Expand the square

2. Split into two fractions Write it as:

So the integrand becomes:

3. Spot the derivative Consider $f(x) = \frac{1}{1+x^2}$. Its derivative is:

That’s exactly the second term, but with a minus sign. So:

Because $f'(x) = -\frac{2x}{(1+x^2)^2}$, so $f(x) + f'(x) = \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2}$.

4. Apply the standard result Hence: …

Method: Expand, then match the $e^x[f+f']$ pattern

For $e^x$ times a squared or compound rational expression, expand first, then split it into $f(x)+f'(x)$.

Steps

Step 1: Expand the bracket.

Multiply out any square so the rational factor is a single fraction, e.g. $\left(\dfrac{1-x}{1+x^2}\right)^2=\dfrac{1-2x+x^2}{(1+x^2)^2}$.

Step 2: Split into $f+f'$. …

Common Mistakes

Mistake 1: Attempting substitution or parts on the squared fraction directly.

Why it's wrong: it produces messy algebra. Correct approach: expand and recognise $\frac{1}{1+x^2}-\frac{2x}{(1+x^2)^2}=f+f'$ with $f=\frac{1}{1+x^2}$.

Mistake 2: Choosing an answer with $(1+x^2)^2$ in the denominator (option C/D). …