∫ ((x² + 1)eˣ / (x + 1)²) dx = f(x)eˣ + C, Where C is constant, then d³f/dx³ at x = 1 is equal to :

Q: Why can't I just use Integration by Parts? It usually works for $e^x$ times a polynomial!

You absolutely can, but it is a trap! Using Integration by Parts here will lead you into a recursive loop or a very long, tedious process that increases the risk of a sign error. The identity $\int e^x [g(x) + g'(x)] dx$ is the 'JEE shortcut' designed to test your pattern recognition, not your stamina. Always look for the $g(x) + g'(x)$ structure first.

Q: How did you know to split $x^2+1$ into $(x^2-1)+2$? Is there a rule for this?

It comes from observing the denominator. We have $(x+1)^2$ in the denominator. We want to create a term that, when differentiated, produces something related to that denominator. By creating $(x^2-1)$, we get $(x-1)(x+1)$, which cancels one $(x+1)$ term. It is not a rule, but a 'geometric' intuition of the expression's structure. You are looking for a way to make the numerator 'talk' to the denominator.

Q: Why do we rewrite $f(x)$ as $1 - 2(x+1)^{-1}$? Isn't the quotient rule safer?

The quotient rule is safe, but it is computationally expensive. For the first derivative, it is fine. For the third derivative, you would be applying the quotient rule repeatedly, which is a breeding ground for silly mistakes. Rewriting it allows you to use the power rule, which is much faster and significantly less prone to error. In JEE, efficiency is just as important as accuracy.

Question

$\int \frac{(x^2+1)e^x}{(x+1)^2} dx = f(x)e^x + C$, Where $C$ is constant, then $\frac{d^3 f}{dx^3}$ at $x=1$ is equal to :

Options:

(A) $-\frac{3}{4}$

(B) $\frac{3}{4}$

(C) $-\frac{3}{2}$

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

### The Art of Pattern Recognition: Mastering the Exponential Integral

Welcome, JEE aspirants. Today, we are not just solving an integral; we are peeling back the layers of a classic problem that tests your ability to see beyond the surface.

When you look at the integral $\int \frac{(x^2+1)e^x}{(x+1)^2} dx$, what do you see? A mess of algebra? A candidate for Integration by Parts? If you see either, take a breath. You are missing the hidden beauty.

### Phase 1: The Golden Identity

In the JEE Advanced arena, whenever you encounter an integral involving $e^x$ multiplied by a rational function, your brain should immediately trigger a specific pattern: the identity $\int e^x [g(x) + g'(x)] dx = e^x g(x) + C$.

This is the **'Golden Identity'** of exponential integration. It is elegant, it is powerful, and it is almost certainly the key to this problem. Our mission is not to integrate directly, but to mold our expression into this exact form.

### Phase 2: Algebraic Surgery

Now, let us look at our integrand: $\frac{x^2+1}{(x+1)^2}$. We need to split this into a function $g(x)$ and its derivative $g'(x)$. The denominator is $(x+1)^2$, which is a huge hint that we need the numerator to contain terms that can cancel out with this denominator.

We perform a bit of 'algebraic surgery.' We rewrite the numerator $x^2+1$ as $(x^2-1) + 2$. Because $x^2-1$ is a difference of squares, we can write it as $(x-1)(x+1)$.

Watch what happens when we split the fraction:

$$
\frac{x^2-1}{(x+1)^2} + \frac{2}{(x+1)^2} = \frac{(x-1)(x+1)}{(x+1)^2} + \frac{2}{(x+1)^2} = \frac{x-1}{x+1} + \frac{2}{(x+1)^2}
$$

Suddenly, the fog clears. If we let $g(x) = \frac{x-1}{x+1}$, let us check its derivative using the quotient rule:

$$
g'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}
$$

It is a perfect match! Our integral is simply $\int e^x [g(x) + g'(x)] dx$, which evaluates to $e^x g(x) + C$. Thus, we have identified $f(x) = \frac{x-1}{x+1}$.

### Phase 3: The Differentiation Trap

We are not done yet. The problem asks for the third derivative of $f(x)$ at $x=1$. Many students will immediately reach for the quotient rule, but applying the quotient rule three times is a recipe for disaster—a single sign error will ruin your entire effort.

Instead, let us rewrite $f(x)$ to make it 'differentiation-friendly':

$$
f(x) = \frac{x+1-2}{x+1} = 1 - 2(x+1)^{-1}
$$

Now, differentiation becomes a simple application of the power rule.

First derivative:
$$
f'(x) = 0 - 2(-1)(x+1)^{-2} = 2(x+1)^{-2}
$$

Second derivative:
$$
f''(x) = 2(-2)(x+1)^{-3} = -4(x+1)^{-3}
$$

Third derivative:
$$
f'''(x) = -4(-3)(x+1)^{-4} = 12(x+1)^{-4} = \frac{12}{(x+1)^4}
$$

### The Final Victory

We have arrived at the finish line. We need to evaluate this at $x=1$:

$$
f'''(1) = \frac{12}{(1+1)^4} = \frac{12}{2^4} = \frac{12}{16} = \mathbf{\frac{3}{4}}
$$

Look at that. By avoiding the brute-force quotient rule and using algebraic insight, we turned a potentially terrifying calculation into a smooth, elegant derivation.

This is the essence of JEE Advanced mathematics: it is not about how hard you work, but how smartly you see the structure of the problem. Keep practicing this level of observation, and you will find that even the most complex problems start to reveal their solutions before you even pick up your pen.

Animated Solution for Mathematics - Indefinite Integration: $\int \frac{( x ^{2} + 1 ) e ^{x}}{( x + 1 ) ^{2}} d x = f (x) e^{x} + C$ , Where $C$ is constant, then $\frac{d ^{3} f}{d x ^{3}}$ at $x = 1$ is equal to :

Select Answer:

Visualized Solution

Problem Analysis

The Special Integral Form

Manipulating the Numerator

Splitting the Fraction

Verifying the Derivative

Identifying $f (x)$

Preparing for Differentiation

First Derivative $f^{'} (x)$

Second Derivative $f^{''} (x)$

Third Derivative $f^{'''} (x)$

Final Calculation at $x = 1$

The Sigma Insight: Evaluation of Special Integral Forms

The Art of Pattern Recognition

Phase 1

Phase 2

Phase 3

The Final Victory

Similar Questions

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫(x+1)2(x2+1)ex​dx=f(x)ex+C, Where C is constant, then dx3d3f​ at x=1 is equal to :

Select Answer:

Visualized Solution

Problem Analysis

The Special Integral Form

Manipulating the Numerator

Splitting the Fraction

Verifying the Derivative

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Preparing for Differentiation

First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Third Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′′(x)

Final Calculation at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1