Problem 83 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{\sqrt{x}} d x\) is equal to (A) \(12\left(\frac{(1+\sqrt[4]{x})^{7 / 3}}{7}+\frac{(1+\sqrt[4]{x})^{4 / 3}}{4}\right)+C\) (B) \(12\left(\frac{(1+\sqrt[4]{x})^{7 / 3}}{7}-\frac{(1+\sqrt[4]{x})^{4 / 3}}{4}\right)+C\) (C) \(6\left(\frac{(1+\sqrt[4]{x})^{7 / 3}}{7}-\frac{(1+\sqrt[4]{x})^{4 / 3}}{4}\right)+C\) (D) none of these

Step-by-Step Solution

Answer

(C) is the correct solution.

Step 1: Identify the substitution

To solve the integral, let's use the substitution method. We start by letting \( u = 1 + \sqrt[4]{x} \). This simplifies the expression inside the integral. The derivative of \( x \) with respect to \( u \) can be obtained by first finding \( \sqrt[4]{x} = u - 1 \), then \( x = (u - 1)^4 \). Next, part of the derivative \( dx \) is calculated.

Step 2: Calculate the derivative of substitution with respect to x

We have \( \sqrt[4]{x} = u - 1 \) which leads to \( x = (u - 1)^4 \). Differentiating this gives \( \frac{dx}{du} = 4(u - 1)^3 \frac{d}{du}(u) = 4(u - 1)^3 \), thus \( dx = 4(u-1)^3 du \).

Step 3: Substitute and simplify the integral

Now substitute into the original integral we have: \[ \int \frac{\sqrt[3]{u}}{\sqrt{(u - 1)^4}} \cdot 4(u-1)^3 \, du \]This simplifies to\[ 4 \int (u^{1/3}) \cdot (u-1)^{2} \, du \] since \( \sqrt{(u-1)^4} = (u-1)^2 \).

Step 4: Expand the expression

Expand the expression \( (u-1)^2 \cdot u^{1/3} \) to simplify the integration:\[ (u-1)^2 = u^2 - 2u + 1 \]So, the integral becomes:\[ 4 \int (u^{1/3})(u^2 - 2u + 1) \, du = 4 \int (u^{7/3} - 2u^{4/3} + u^{1/3}) \, du \], which is a sum of simpler integrals.

Step 5: Integrate each term separately

Integrate each term:- \( \int u^{7/3} \, du = \frac{u^{10/3}}{10/3} = \frac{3}{10}u^{10/3} \)- \( \int -2u^{4/3} \, du = -2 \times \frac{u^{7/3}}{7/3} = -\frac{6}{7}u^{7/3} \)- \( \int u^{1/3} \, du = \frac{u^{4/3}}{4/3} = \frac{3}{4}u^{4/3} \)Combine these results and multiply by 4:

Step 6: Evaluate and simplify the expression

Multiply each part by 4 and simplify:\[ 4 \times \left( \frac{3}{10}u^{10/3} - \frac{6}{7}u^{7/3} + \frac{3}{4}u^{4/3} \right) \]Simplify to get:\[ \frac{12}{10}u^{10/3} - \frac{24}{7}u^{7/3} + 3u^{4/3} \]Final simplification gives:\[ \frac{6}{5}u^{10/3} - \frac{24}{7}u^{7/3} + 3u^{4/3} + C \].

Step 7: Re-substitute and compare with options

Substitute back \( u = 1 + \sqrt[4]{x} \) into the simplified integrals and compare the expression with the given options. After re-substitution, you should verify which option corresponds to this final integration and simplification.

Key Concepts

Substitution Method

The substitution method is a technique used to simplify integrals by introducing a new variable. It is particularly useful when dealing with complex expressions that contain nested functions. In the given exercise, we start by identifying a part of the integrand that can be substituted to make integration more straightforward.

Here, we chose to let \( u = 1 + \sqrt[4]{x} \), transforming the original integral into a simpler form in terms of \( u \). This change of variables requires calculating the differential \( dx \) in terms of \( du \). We do this by differentiating \( x = (u - 1)^4 \), which leads to \( dx = 4(u-1)^3 \, du \).

Substituting back into the integral allows us to express the problem entirely in terms of \( u \). This simplifies the integration process significantly as it avoids dealing directly with the more complicated initial expression. With practice, recognizing appropriate substitution becomes easier and is a powerful tool in simplifying integration.

Integration by Parts

Although the original problem might not directly require integration by parts, understanding this technique is crucial for tackling various integral situations. Integration by parts is particularly useful when dealing with products of functions, such as \( u(x) \cdot v'(x) \).

The formula for integration by parts is derived from the product rule for differentiation: \[ \int u \, dv = uv - \int v \, du \]

The goal is to select \( u \) and \( dv \) wisely such that the resulting integrals are simpler to evaluate. Typically, we choose \( u \) to be a function that becomes simpler when differentiated, and \( dv \) to be a function whose integral is easily found.

While substitution was the method of choice in the original exercise, being proficient in integration by parts provides a broader toolkit for solving complex integrals, especially when substitution alone is insufficient.

Integration Techniques

In calculus, mastering integration involves learning various techniques to handle different types of integrals. The substitution method and integration by parts are just two of the many techniques available. Each technique is suited to different forms of integrals, and often a combination might be necessary.

- **Substitution**: Useful for simplifying integrands by changing variables, especially when dealing with nested functions. It can sometimes transform a problem into a basic integral form, making it straightforward to solve. - **Integration by Parts**: Ideal for products of functions, drawing on the derivative product rule. - **Partial Fractions**: A valuable technique when dealing with rational functions, breaking them down into simpler, more manageable pieces. - **Trigonometric Integrals**: Applies specific identities to simplify the integration of trigonometric functions.

Choosing the correct technique often comes with experience and practice. A thorough understanding of each allows for more flexibility and confidence in handling a wide range of integration problems.