Problem 82 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int \frac{\sqrt{1+\sqrt{x}}}{x} d x\) is equal to (A) \(2 \sqrt{1+\sqrt{x}}-2 \log \left(\frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1}\right)+C\) (B) \(4 \sqrt{1+\sqrt{x}}+2 \log \left(\frac{\sqrt{1+\sqrt{x}}+1}{\sqrt{1+\sqrt{x}}-1}\right)+C\) (C) \(4 \sqrt{1+\sqrt{x}}+2 \log \left(\frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}+1}}\right)+C\) (D) none of these

Step-by-Step Solution

Answer

The integral is option (B).

Step 1: Choose a suitable substitution

Given the integral \( \int \frac{\sqrt{1+\sqrt{x}}}{x} \, dx \), we can use the substitution \( u = \sqrt{x} \), which implies that \( x = u^2 \) and \( dx = 2u \, du \). The integral becomes \( \int \frac{\sqrt{1+u}}{u^2} \cdot 2u \, du = 2 \int \frac{\sqrt{1+u}}{u} \, du \).

Step 2: Simplify the integral

The integral now is \( 2 \int \frac{\sqrt{1+u}}{u} \, du \). To simplify, set \( v = \sqrt{1+u} \). Then, \( v^2 = 1 + u \) and \( u = v^2 - 1 \), so \( du = 2v \, dv \). Thus, the integral becomes \( 2 \int \frac{v}{v^2 - 1} \, 2v \, dv = 4 \int \frac{v^2}{v^2 - 1} \, dv \).

Step 3: Split and integrate

Rewriting \( \frac{v^2}{v^2 - 1} \) as \( 1 + \frac{1}{v^2 - 1} \), the integral becomes \( 4 \left( \int 1 \, dv + \int \frac{1}{v^2-1} \, dv \right) \). The first integral \( \int 1 \, dv \) is \( v \). The second integral involves partial fractions: \( \frac{1}{v^2 - 1} = \frac{1}{2} \left( \frac{1}{v-1} - \frac{1}{v+1} \right) \).

Step 4: Integrate using partial fractions

The integral of \( \frac{1}{v^2-1} \) becomes \( \int \frac{1}{2} \left( \frac{1}{v-1} - \frac{1}{v+1} \right) \, dv = \frac{1}{2} \left( \log|v-1| - \log|v+1| \right) = \frac{1}{2} \log \left| \frac{v-1}{v+1} \right| \). Thus the integral \( \int \frac{1}{v^2-1} \, dv \) is \( \frac{1}{2} \log \left| \frac{v-1}{v+1} \right| \).

Step 5: Combine and simplify the results

Combining the results from Steps 3 and 4, the entire integral expression is \( 4(v) + 2 \log \left| \frac{v-1}{v+1} \right| + C \). Substitute back for \( v = \sqrt{1 + u} \) and \( u = \sqrt{x} \) to get \( 4\sqrt{1+\sqrt{x}} + 2 \log \left| \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right| + C \).

Step 6: Verify expression with given options

Now compare the expression obtained \( 4\sqrt{1+\sqrt{x}} + 2 \log \left| \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right| + C \) with the given options. It matches option (B).

Key Concepts

Substitution Method

The substitution method is often a great first step when dealing with complex integrals. It helps to simplify the integral by changing variables.

In the original exercise \(\int \frac{\sqrt{1+\sqrt{x}}}{x} \, dx\), the form of the expression suggests using substitution.
We choose \(u = \sqrt{x}\) because \(\sqrt{1+\sqrt{x}}\) is easier to handle as \(1+u\).
This substitution simplifies the derivative \(x\) features as well: \(x = u^2\) and \( dx = 2u \, du\).

This transformation simplifies the integral dramatically:

• Always identify parts of the integral that seem to complicate the integration.
• Choose substitutions based on expressions that simplify when differentiated, like powers or square roots.
• Remember to express \(dx\) in terms of the new variable, which is crucial for seamless integration.

By smart substitutions, you transform an unwieldy expression to a manageable one, setting the stage for further techniques like partial fractions.

Partial Fraction Decomposition

Partial fraction decomposition is a valuable algebraic technique.
It’s used to break down complex rational expressions into simpler parts that can be easily integrated.
In this exercise, it came into play after substitution simplified the initial integral.

We arrived at the integral \(\int \frac{v^2}{v^2 - 1} \, dv\).
Partial fraction decomposition lets us rewrite this as two simpler fractions: \(1 + \frac{1}{v^2 - 1}\).
Here, \(\frac{1}{v^2-1}\) is further decomposed:

• Expressed as \(\frac{1}{2} \left( \frac{1}{v-1} - \frac{1}{v+1} \right)\).
• This step hinges on recognizing \(v^2-1\) as a difference of squares: \(v^2 - 1 = (v - 1)(v + 1)\).
• The decomposition simplifies the integration process by transforming complex expressions into basic logarithmic forms.

Partial fractions are particularly useful for rational functions.
They break down polynomial ratios into sums of fractions, making integration more straightforward.

Indefinite Integrals

Indefinite integrals represent a broad class of problems, where you're not provided a boundary for integration.
This means the solution includes an arbitrary constant, \(C\), because the derivative of a constant is zero.

In the exercise, after employing substitution and partial fractions, we arrived at an indefinite integral from: \(\int 1 \, dv + \int \frac{1}{v^2-1} \, dv\).
The indefinite integration resulted in:

• First term integrates to \(v\), derived simply from \(\int 1 \, dv\).
• Second term needed partial fraction decomposition and returned a logarithmic form.
• Ultimately, where possible solutions integrated indefinitely feature the \(C\), indicating they are solutions to a family of functions.

Indefinite integrals form the foundation for numerous applications in physics, engineering, and more.
They allow us to understand solutions generally, before setting specific limits or conditions.