The integral ∫ (2x³ - 1)/(x⁴ + x) dx is equal to : (A) 1/2 log_e |x³ + 1|/x² + C (B) 1/2 log_e (x³ +

Q: Why did we choose to divide by $x^2$ specifically? Why not $x$ or $x^3$?

Great question! We choose the power of $x$ that makes the denominator's derivative appear in the numerator. By dividing by $x^2$, we transform the denominator into $x^2 + \frac{1}{x}$, whose derivative is $2x - \frac{1}{x^2}$. This perfectly matches the modified numerator. It is all about creating a 'perfect pair' for substitution.

Q: I got $\ln|\frac{x^3+1}{x}| + C$, but the options have a $\frac{1}{2}$ in front. Did I do something wrong?

Not at all! You are mathematically correct. The options are simply written in an equivalent form. Remember the log property $n \ln(a) = \ln(a^n)$. Here, $\ln|y| = \frac{1}{2} \ln(y^2)$. By squaring the argument, the options introduce that $\frac{1}{2}$ factor. It is just a different 'outfit' for the same answer.

Q: Can I use partial fractions here instead of substitution?

Technically, yes, but it would be a nightmare! You would have to factor $x^4 + x$ into $x(x+1)(x^2 - x + 1)$. Setting up the partial fraction decomposition would involve four constants and a massive system of equations. Substitution is the 'elegant' path—always look for the shortcut before diving into the long algebra.

Q: How do I know when to use substitution versus partial fractions?

Think of it as a hierarchy. First, check if the numerator is the derivative (or a multiple) of the denominator. If yes, use substitution. If the denominator is factorable and the numerator is of lower degree, partial fractions are your friend. If the degree of the numerator is higher, perform long division first!

Q: What happens to the constant $C$ when we manipulate the log?

The constant $C$ is an arbitrary 'catch-all' for any missing terms. When we perform algebraic manipulations on the log, the constant remains $C$ because any constant added or multiplied still results in an arbitrary constant. It is the 'guardian' of the indefinite integral!

Question

The integral $\int \frac{2x^3 - 1}{x^4 + x} dx$ is equal to :

Options:

(A) $\frac{1}{2} \log_e \frac{|x^3 + 1|}{x^2} + C$

(B) $\frac{1}{2} \log_e \frac{(x^3 + 1)^2}{|x^3|} + C$

(C) $\frac{1}{2} \log_e \frac{|x^3 + 1|}{x^3} + C$

(D) $\frac{1}{2} \log_e \frac{|x^3 + 1|}{x^2} + C$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of powers.

We are looking at the integral:

$$I = \int \frac{2x^3 - 1}{x^4 + x} dx$$

When you see a rational function where the degree of the denominator is only one higher than the numerator, your intuition should immediately scream "substitution!"

However, if you set $u = x^4 + x$, the derivative is $4x^3 + 1$. Since our numerator is $2x^3 - 1$, they are not a direct match. This is where the true test of an engineer begins: not by brute force, but by strategic manipulation.

### The Strategic Pivot: Dividing by $x^2$

Imagine you are a sculptor. To reveal the "statue" hidden within the denominator, we need to perform a clever transformation.

Let us divide both the numerator and the denominator by $x^2$. This is the perfect bridge to connect the $x^3$ term in the numerator to the $x^4$ term in the denominator.

When we divide the numerator, we get:

$$\frac{2x^3}{x^2} - \frac{1}{x^2} = 2x - \frac{1}{x^2}$$

When we divide the denominator, we get:

$$\frac{x^4}{x^2} + \frac{x}{x^2} = x^2 + \frac{1}{x}$$

Suddenly, the integral transforms into:

$$I = \int \frac{2x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} dx$$

Do you see it now? The numerator is exactly the derivative of the denominator. This is the "Aha!" moment that separates the novice from the master.

### The Elegant Substitution

Let us define our new variable $t = x^2 + \frac{1}{x}$. Now, let us find the differential $dt$.

Differentiating with respect to $x$, we get:

$$\frac{dt}{dx} = 2x - \frac{1}{x^2} \Rightarrow dt = \left( 2x - \frac{1}{x^2} \right) dx$$

The entire numerator of our integral is swallowed up by $dt$. The integral simplifies to the most beautiful form in all of calculus:

$$I = \int \frac{1}{t} dt = \ln|t| + C$$

### The Final Polish: Matching the Form

We must now return to the world of $x$. Substituting back, we get:

$$I = \ln\left| x^2 + \frac{1}{x} \right| + C$$

Simplifying the fraction inside the logarithm, we have:

$$I = \ln\left| \frac{x^3 + 1}{x} \right| + C$$

If the required options contain a $\frac{1}{2}$ factor, we use the logarithmic identity $\ln(y) = \frac{1}{2} \ln(y^2)$. Applying this, we rewrite our answer as:

$$I = \frac{1}{2} \ln\left( \frac{x^3 + 1}{x} \right)^2 + C = \mathbf{\frac{1}{2} \ln \frac{(x^3 + 1)^2}{x^2} + C}$$

This is the beauty of JEE mathematics. It is not just about finding the answer; it is about recognizing the different forms of truth. You have navigated the complexity, performed the substitution, and matched the form.

Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{2 x ^{3} - 1}{x ^{4} + x} d x$ is equal to :

Select Answer:

Visualized Solution

Analyze the Integral Structure

Strategic Division by $x^{2}$

Identify the Substitution $t = x^{2} + \frac{1}{x}$

Calculate the Differential $d t$

Perform Substitution & Integrate

Back-Substitution to $x$

Final Result and Form Matching

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Strategic Pivot

The Elegant Substitution

The Final Polish

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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; } The integral ∫x4+x2x3−1​dx is equal to :

Select Answer:

Visualized Solution

Analyze the Integral Structure

Strategic Division by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

Identify the Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=x2+x1​

Calculate the Differential .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dt

Perform Substitution & Integrate

Back-Substitution to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x