Problem 5 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\int \frac{(\sqrt{x})^{5}}{(\sqrt{x})^{7}+x^{6}} d x=a \operatorname{In}\left(\frac{x^{k}}{x^{k}+1}\right)+c\), the values of \(a\) and \(k\) respectively are (A) \(\frac{5}{2}\) and \(\frac{2}{5}\) (B) \(\frac{2}{5}\) and \(\frac{5}{2}\) (C) \(\frac{5}{2}\) and 2 (D) none of there

Step-by-Step Solution

Answer

The values of \(a\) and \(k\) are \(\frac{2}{5}\) and \(\frac{5}{2}\) respectively.

Step 1: Simplify the Integrand

Rewrite the given integral \( \int \frac{(\sqrt{x})^5}{(\sqrt{x})^7 + x^{6}} \, dx \) by expressing powers of \( \sqrt{x} \) in terms of \( x \). We have \( (\sqrt{x})^5 = x^{5/2} \) and \( (\sqrt{x})^7 = x^{7/2} \). Then rewrite the integral as \( \int \frac{x^{5/2}}{x^{7/2} + x^{6}} \, dx \).

Step 2: Simplify the Denominator Further

Factor out the lowest power of \( x \) in the denominator. Notice that \( x^{6} = x^{12/2} \), so in the expression \( x^{7/2} + x^{6} \), the lowest power is \( x^{7/2} \). Hence, factor to get \( x^{7/2}(1 + x^{5/2}) \), and rewrite the integrand as \( \frac{x^{5/2}}{x^{7/2}(1 + x^{5/2})} = \frac{x^{-1}}{1 + x^{5/2}} \).

Step 3: Substitution

Let \( u = x^{5/2} \). Then \( du = \frac{5}{2}x^{3/2} \, dx \) or equivalently \( dx = \frac{2}{5}x^{-3/2} \, du \). Using this substitution, change the integral to \( \int \frac{1}{u + 1} \cdot \frac{2}{5} \, du \).

Step 4: Solve the Integral

The integral \( \int \frac{2}{5} \cdot \frac{1}{u + 1} \, du \) is straightforward. It evaluates to \( \frac{2}{5} \ln |u+1| + C \).

Step 5: Back Substitute

Substitute back \( u = x^{5/2} \) into the integrated function to obtain \( \frac{2}{5} \ln |x^{5/2} + 1| + C \).

Step 6: Identify \( a \) and \( k \)

From the expression \( a \ln \left( \frac{x^k}{x^k + 1} \right) + c \), we identify that \( a = \frac{2}{5} \) and \( x^k = x^{5/2} \), thus \( k = \frac{5}{2} \).

Key Concepts

Integration Techniques

Integration Techniques are powerful tools for solving complex mathematical problems that involve finding the area under a curve or the accumulation of quantities. These techniques can vary based on the complexity and the form of the integrand. Here are some common techniques:

• Basic Integration: This involves directly applying the fundamental rules of integration to evaluate simple integrals.
• Substitution Method: Often used when the integrand can be simplified by substituting a part of it with a single variable, simplifying the expression into a form that's easier to integrate.
• Integration by Parts: Useful for integrals that are products of functions, utilizing the relationship between the integral of a product and the product of their derivatives.
• Partial Fraction Decomposition: Applied to rational functions, breaking them into simpler fractions that are easier to integrate.

Understanding which technique to apply is crucial. For instance, if faced with a complex function that doesn't fit basic integration rules, looking for substitution possibilities could simplify the process. These techniques enhance problem-solving skills by allowing us to handle a variety of integrals efficiently.

Substitution Method

The Substitution Method is like a clever trick in integration. It's used to simplify integrals by changing variables. When you see a complicated function, substitution can often turn it into something much simpler.
Here's how it works:

• Identify a Substitution: Look for a part of the integrand that, if replaced by a single variable, would simplify the entire expression. This often involves setting a variable like \( u \) equal to a function inside the integral.
• Calculate the Differential: Once you've set \( u \), differentiate it to find \( du \). This helps in expressing \( dx \) in terms of \( du \).
• Substitute and Simplify: Replace all instances of the original variable and \( dx \) in the integral with the new variable and \( du \). This often results in a simpler integral that's easier to solve.
• Integrate: Solve the new, simpler integral in terms of \( u \).
• Back Substitute: Finally, replace \( u \) with the original expression from your substitution to return to the original variable.

This method not only simplifies the computation but also enhances understanding of the relationship between different parts of the integrand. In our exercise, substituting \( u = x^{5/2} \) led to a much more straightforward integral to solve.

Mathematical Problem Solving

Mathematical Problem Solving is a fundamental skill in mathematics that goes beyond just performing computations. It involves understanding problems, using creativity, and applying logical reasoning to find solutions.
In the context of calculus and integration:

• Understanding the Problem: Before you start solving an integral, it's crucial to know what the problem is asking for. Here, understanding that we need to determine constants from a given integral expression is key.
• Identifying Techniques: Figuring out which integration techniques to use can save time and avoid errors. In our example, recognizing that substitution could simplify the process was essential.
• Breaking Down the Problem: Decompose the integral into simpler parts. This might involve rewriting complex expressions or factoring, as we did by simplifying the denominator.
• Solution Verification: Once a solution is found, it's important to check it against the problem to ensure that it meets all the requirements. For this exercise, verifying the expressions for \( a \) and \( k \) ensured that the solution was correct.

By honing these skills, students not only solve integrals more efficiently but also gain a deeper appreciation of the interconnected nature of math concepts. This process not only finds the answer but strengthens analytical thinking and problem-solving confidence.