Problem 33 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\int \frac{\sqrt{1+\sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x\) is equal to (A) \(\left(1+x^{1 / 3}\right)^{3 / 2}+C\) (B) \(-\left(1+x^{1 / 3}\right)^{1 / 2}+C\) (C) \(2\left(1+x^{1 / 3}\right)^{3 / 2}+C\) (D) none of these

Step-by-Step Solution

Answer

(C)

Step 1: Simplify the integrand

We have the integral \( I = \int \frac{\sqrt{1 + \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} \, dx \). Notice that \( \sqrt[3]{x^2} = x^{2/3} \), so the integrand becomes \( \frac{\sqrt{1 + x^{1/3}}}{x^{2/3}} \). We can rewrite this as \( (1 + x^{1/3})^{1/2} \cdot x^{-2/3} \).

Step 2: Use substitution to simplify

Let \( u = 1 + x^{1/3} \). Then \( du = \frac{1}{3}x^{-2/3} \, dx \), which implies \( 3 \, du = x^{-2/3} \, dx \). Substituting in terms of \( u \), the integral becomes \( 3 \int u^{1/2} \, du \).

Step 3: Integrate with respect to u

Integrate \( 3 \int u^{1/2} \, du \). The antiderivative of \( u^{1/2} \) is \( \frac{2}{3}u^{3/2} \), so the antiderivative is \( 3 \cdot \frac{2}{3}u^{3/2} = 2u^{3/2} \).

Step 4: Substitute back for x

Recall that \( u = 1 + x^{1/3} \). Substitute back to get \( 2(1 + x^{1/3})^{3/2} \). Thus, the integral evaluates to \( 2(1 + x^{1/3})^{3/2} + C \).

Step 5: Select the matching option

Comparing our solution to the given options, we find that \( 2(1 + x^{1/3})^{3/2} + C \) matches option (C).

Key Concepts

Substitution Method

The substitution method is a crucial tool for solving challenging integrals in calculus. It involves changing the variable of integration to simplify the expression. Often referred to as "u-substitution", this approach helps transform a complex integral into a more manageable form.

When using substitution, the first step is to identify a part of the integrand that can be replaced with a single variable, usually denoted as \( u \). The goal is to choose a substitution that simplifies the integrand significantly. In our example, we have chosen \( u = 1 + x^{1/3} \) for the expression \( \sqrt{1 + \sqrt[3]{x}} \).

After deciding on \( u \), differentiate it with respect to \( x \) to find \( du \). Here, differentiating gives us \( du = \frac{1}{3}x^{-2/3} \, dx \). By rearranging, we can express \( dx \) as a function of \( du \), allowing us to substitute the entire integral in terms of \( u \). This change simplifies the integration process, reducing our problem to finding the antiderivative of a simpler function in \( u \).

• Identify and define the substitution: \( u = 1 + x^{1/3} \).
• Differentiation to find \( du \): \( du = \frac{1}{3}x^{-2/3} \, dx \).
• Express \( dx \) in terms of \( du \).
• Substitute in the original integral to simplify.

Antiderivatives

Antiderivatives are the reverse operation of derivatives. They are used to find a function whose derivative is the given function. In integral calculus, antiderivatives are directly linked to finding integrals.

When dealing with an integral, especially after substitution, the next step is to integrate with respect to the new variable. In our problem, after substituting \( u \), the integral reduces to \( 3 \int u^{1/2} \, du \). The process here involves finding the antiderivative of \( u^{1/2} \), which is essentially the function whose derivative would be \( u^{1/2} \).

To compute this, remember the power rule for antiderivatives: the integral of \( u^n \) with respect to \( u \) is \( \frac{u^{n+1}}{n+1} \). Applying this rule, we find that the antiderivative of \( u^{1/2} \) is \( \frac{2}{3}u^{3/2} \). Hence, multiplying by 3 as per the expression, gives us \( 2u^{3/2} \). This result must be integrated back into the original variable \( x \) to find the solution to the initial problem.

• Antiderivatives undo derivatives to find original functions.
• Use the power rule to find the antiderivative of \( u^{1/2} \).
• Multiply by any constants from substitutions.
• Convert back to original variable for final answer.

Integral Calculus

Integral calculus is a fundamental branch of mathematics focusing on accumulation and areas. It works as the inverse operation to differential calculus and can calculate quantities like areas under curves.

In the given problem, integral calculus allows us to transition from a compounded expression to a straightforward solution. Integration, often requiring techniques like substitution, is pivotal here because it translates complex integrands into solvable forms.

The process of integration doesn't just solve equations; it also provides insights into the behavior of functions. By understanding and applying various techniques, such as the substitution method used here, we gain powerful tools to analyze and interpret mathematical models.

• Integral calculus finds cumulative quantities like distances and areas.
• Acts as the inverse of differentiation, a key part of finding solutions.
• Use techniques like substitution for transforming integrals.
• Integral solutions reveal insights into mathematical functions and models.