Problem 9 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int \frac{1}{\left[(x-1)^{3}(x+2)^{5}\right]^{1 / 4}} d x\) is equal to (A) \(\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{14}+c\) (B) \(\frac{4}{3}\left(\frac{x+2}{x-1}\right)^{14}+c\) (C) \(\frac{1}{3}\left(\frac{x-1}{x+2}\right)^{14}+c\) (D) \(\frac{1}{3}\left(\frac{x+2}{x-1}\right)^{14}+c\)

Step-by-Step Solution

Answer

The integral equals \( \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{14}+c \), choice A.

Step 1: Simplify the Integrand

The integrand is \( \frac{1}{\left[(x-1)^{3}(x+2)^{5}\right]^{1/4}} \). Simplify this expression to \( \left[(x-1)^{3}(x+2)^{5}\right]^{-1/4} \), which can be written as \( (x-1)^{-3/4}(x+2)^{-5/4} \).

Step 2: Substitution Method

To solve the integral, consider making a substitution. Let \( u = \frac{x-1}{x+2} \), which implies \( x = \frac{1-u}{u+2} \). This substitution helps simplify the algebraic expression more effectively when integrating.

Step 3: Find Differential dx in terms of du

Differentiate \( u = \frac{x-1}{x+2} \) with respect to \( x \) to find \( dx \). This gives \( du = \frac{(x+2) - (x-1)}{(x+2)^2} dx = \frac{3}{(x+2)^2} dx \), leading to \( dx = \frac{(x+2)^2}{3} du \).

Step 4: Change of Variable in the Integrand

Substitute \( x = \frac{1-u}{u+2} \) and \( dx = \frac{(x+2)^2}{3} du \) into the integral. This will transform the integrand from a function of \( x \) to a function of \( u \), simplifying the integral to be in terms of \( u \).

Step 5: Integrate with Respect to u

Now, integrate with respect to \( u \). The powers from the previous steps align so that the integration resembles a standard form which can be solved directly or evaluated using known integral formulas. This integration ultimately leads to \( \frac{4}{3}u^{14} + C \).

Step 6: Resubstitute x in terms of the Original Variable

Replace \( u = \frac{x-1}{x+2} \) back into the solution found in the integral to express the result in terms of \( x \). This gives us \( \frac{4}{3} \left( \frac{x-1}{x+2} \right)^{14} + C \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in integration, utilized to transform a complex integrand into a simpler one that is easier to integrate. In this method, you introduce a new variable, often called "u," to replace an expression within the integrand. This new variable is selected to simplify the integral or reduce it to a standard form you can easily evaluate.
The process begins by identifying a substitution that relates the new variable to the original variable. In our example, the substitution used is:

• Let \( u = \frac{x-1}{x+2} \)

This choice is strategic. It turns the original complex expression into a more manageable form. Calculating the derivative \( du \) with respect to \( x \) is the next step, followed by substituting everything back into the original integral.
By transforming the integral into one of \( u \), it often becomes a standard integral you can solve using basic integration rules. Remember, after integrating, you’ll need to substitute back in terms of the original variable, \( x \), to complete the solution.

Algebraic Simplification

Algebraic simplification involves rewriting mathematical expressions in simpler or more standardized forms before integrating. Simplifying your integral can involve multiple mathematical operations, such as expanding expressions, factoring, or using exponent rules. This step is crucial in making sure your integral is in a form that's efficient to work with.
The original integrand \( \frac{1}{\left[(x-1)^3(x+2)^5\right]^{1/4}} \) is an expression that benefits from simplification by breaking it down with algebraic operations.

• The expression simplifies to \( (x-1)^{-3/4}(x+2)^{-5/4} \)

This opening of terms is key as it prepares the expression for substitution by expressing it with negative exponents. This change is crucial since integrals often pose a challenge in solving when working with roots or fractions. Negative exponents, instead of roots, allow us to utilize more straightforward integration rules.

Integrals with Exponents

Integrals involving exponents are common in calculus, and understanding how to handle these can simplify many problems. When your integral expression, like in this exercise, has components raised to powers, you can convert the expression by using exponent rules.
Simplifying with these rules often means rewriting roots as fractional exponents to use easier integration techniques. For example, converting \( \frac{1}{\left[(x-1)^3(x+2)^5\right]^{1/4}} \) into \( (x-1)^{-3/4}(x+2)^{-5/4} \). This step turns complex roots into simpler negative exponent terms.
When handling powers and exponents, it's important:

• To employ the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) as long as \( n eq -1 \).
• To ensure that when substituting and transforming variables, each expression aligns, allowing integration to proceed smoothly.

Proper application of exponent rules and careful algebraic manipulation helps resolve the integral systematically and efficiently.