The integral ∫ (1 + x - 1/x) e^(x + 1/x) dx is equal to (a) (x + 1) e^(x + 1/x) + c (b) -x e^(x + 1/

Q: Wait, why do we split the integral? Can't we just solve it directly?

Great question! While you could potentially try other methods, splitting the integral is a strategic move. By breaking it into $\int e^{x + \frac{1}{x}} dx + \int (x - \frac{1}{x}) e^{x + \frac{1}{x}} dx$, we isolate the part that contains the derivative of the exponent. This makes the Integration by Parts step much more obvious and prevents you from getting lost in the algebra.

Q: How do I know when to use Integration by Parts (IBP) in a problem like this?

Look for the structure of the integrand. If you see a product of two functions where one is an exponential $e^{f(x)}$ and the other is an algebraic expression that involves $f'(x)$, IBP is almost always the right path. In this case, the algebraic part $(1 + x - \frac{1}{x})$ contains the derivative of the exponent $(1 - \frac{1}{x^2})$ once you factor it correctly. Recognizing this pattern is the key to solving these JEE-level problems.

Q: What if I didn't see the derivative of the exponent immediately?

That is perfectly normal! The trick is to always start by examining the exponent. If you see $e^{x + \frac{1}{x}}$, calculate its derivative on the side. Once you see that the derivative is $(1 - \frac{1}{x^2}) e^{x + \frac{1}{x}}$, you can look back at your integrand and try to manipulate the algebraic terms to match that derivative. It is all about pattern matching and persistence.

Question

The integral $\int \left( 1 + x - \frac{1}{x} \right) e^{x + 1/x} dx$ is equal to

Options:

(A) $(x + 1) e^{x + 1/x} + c$

(B) $-x e^{x + 1/x} + c$

(C) $(x - 1) e^{x + 1/x} + c$

(D) $x e^{x + 1/x} + c$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex integral:

$$I = \int \left(1 + x - \frac{1}{x}\right) e^{x + \frac{1}{x}} dx$$

At first glance, it looks intimidating. The combination of algebraic terms and an exponential function with a non-linear exponent can make anyone pause. However, in the world of JEE Advanced, intimidation is just a sign that a beautiful, elegant solution is waiting to be uncovered.

### The Strategic Split

The first step in our journey is to simplify our perspective. We have an integrand that is a product of an algebraic expression and an exponential term. Let us distribute the exponential term to see what we are really dealing with:

$$I = \int e^{x + \frac{1}{x}} dx + \int \left(x - \frac{1}{x}\right) e^{x + \frac{1}{x}} dx$$

By splitting the integral, we have created two distinct paths. The first part remains as is, but the second part, $\int (x - \frac{1}{x}) e^{x + \frac{1}{x}} dx$, holds the secret.

### The Hidden Derivative

Now, let us focus on the exponent: $f(x) = x + \frac{1}{x}$. If we differentiate this with respect to $x$, we get:

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

This is the "Aha!" moment. Notice that if we factor an $x$ out of the algebraic part of our second integral, we get $x(1 - \frac{1}{x^2})$. Suddenly, the derivative of the exponent appears right before our eyes. We can rewrite the second integral as:

$$\int x \left(1 - \frac{1}{x^2}\right) e^{x + \frac{1}{x}} dx$$

### The IBP Dance

With the derivative identified, we are ready for the main event: Integration by Parts. We choose $u = x$, which gives us $du = dx$.

Then, we choose $dv = (1 - \frac{1}{x^2}) e^{x + \frac{1}{x}} dx$. Because we know that the derivative of $e^{x + \frac{1}{x}}$ is exactly $(1 - \frac{1}{x^2}) e^{x + \frac{1}{x}}$, integrating $dv$ is trivial: $v = e^{x + \frac{1}{x}}$.

Applying the formula $\int u \, dv = uv - \int v \, du$, we get:

$$x e^{x + \frac{1}{x}} - \int e^{x + \frac{1}{x}} dx$$

### The Elegant Collapse

Now, look at the full picture. We substitute this result back into our original split integral:

$$I = \int e^{x + \frac{1}{x}} dx + \left[x e^{x + \frac{1}{x}} - \int e^{x + \frac{1}{x}} dx\right] + C$$

The term $\int e^{x + \frac{1}{x}} dx$ appears once as a positive and once as a negative. They cancel each other out perfectly.

This leaves us with the final, clean answer:

**$$I = x e^{x + \frac{1}{x}} + C$$**

This is the beauty of calculus—when you find the right pattern, the complexity simply melts away. Keep practicing, keep looking for those hidden derivatives, and you will master these problems in no time.

Animated Solution for Mathematics - Indefinite Integration: The integral $\int (1 + x - \frac{1}{x}) e^{x + 1/ x} d x$ is equal to

Select Answer:

Visualized Solution

Analyzing the Integral $I$

Splitting the Expression

Factoring the Second Term

Identifying the Derivative

Integration by Parts Setup

Finding the Integral $v$

Applying the IBP Formula

Combining and Cancelling

Final Result

The Sigma Insight: Integration by Parts

Analyzing the Setup

The Strategic Split

The Hidden Derivative

The IBP Dance

The Elegant Collapse

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