Problem 14 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\int \frac{x^{2} d x}{\left(1+x^{2}\right)\left(1+\sqrt{1+x^{2}}\right)}\) and \(f(0)=0\), then the value of \(f(1)\) is (A) \(\log (1+\sqrt{2})\) (B) \(\log (1+\sqrt{2})-\frac{\pi}{4}\) (C) \(\log (1+\sqrt{2})+\frac{\pi}{2}\) (D) none of these

Step-by-Step Solution

Answer

(A) \( \log(1 + \sqrt{2}) \)

Step 1: Simplify the Integral

We need to solve the integral \( f(x) = \int \frac{x^2 \, dx}{(1+x^2)(1 + \sqrt{1 + x^2})} \). This can be approached by using substitution to simplify the integrand. Let us set \( t = \sqrt{1 + x^2} \). Then, \( x^2 = t^2 - 1 \) and \( dx = \frac{t}{x} \, dt \).Now, substitute these values into the integral:\[\int \frac{(t^2 - 1)x^2}{(1+x^2)(1+t)} \, \frac{t}{x} \, dt = \int \frac{(t^2 - 1)t}{t^2(t+1)} \, dt \]\[= \int \frac{t^2 - 1}{t(t+1)} \, dt \].

Step 2: Partial Fraction Decomposition

Using partial fraction decomposition, we rewrite:\[\frac{t^2 - 1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1}\]Solving for \(A\) and \(B\), we equate coefficients:\( t^2 - 1 = A(t+1) + B(t)\).Let \(t = 0\) to find \(A\), giving:\( -1 = A\cdot 1 \) so \( A = -1\).Let \( t = -1 \) to find \( B\), giving:\( 2 = B(-1)\cdot(-1) \) so \( B = 1\).Thus, the decomposition is:\[\frac{-1}{t} + \frac{1}{t+1}\].

Step 3: Integrate Each Term

Now, integrate both terms separately:\[ \int \frac{-1}{t} \, dt = -\log|t| \]\[ \int \frac{1}{t+1} \, dt = \log|t+1| \]Combine the results:\[\int \left(-\frac{1}{t} + \frac{1}{t+1}\right) \, dt = -\log|t| + \log|t+1| \]\[= \log\left|\frac{t+1}{t}\right|\].

Step 4: Substitute Back in Terms of x

Recall that \( t = \sqrt{1 + x^2} \). Substitute back:\[ \log\left|\frac{\sqrt{1 + x^2} + 1}{\sqrt{1 + x^2}}\right| \]\[= \log(1 + \sqrt{1 + x^2}) - \log(\sqrt{1 + x^2})\].

Step 5: Evaluate from 0 to 1

We must evaluate this definite integral from 0 to 1 using the Fundamental Theorem of Calculus:\[ f(1) - f(0) = \left[ \log(1 + \sqrt{1 + x^2}) - \log(\sqrt{1 + x^2}) \right]_0^1 \]First evaluate \( f(0) \):- For \( x = 0 \), \( \log(1 + \sqrt{1 + 0^2}) - \log(\sqrt{1 + 0^2}) = 0 - 0 = 0\).Then evaluate \( f(1) \):- For \( x = 1 \), \( \log(1 + \sqrt{1+1^2}) - \log(\sqrt{1+1^2}) \), which simplifies to \( \log(1 + \sqrt{2}) - \log(\sqrt{2}) \).Thus, \( f(1) = f(0) + \log(1+\sqrt{2}) - \log(\sqrt{2}) \). Given \( f(0) = 0 \), we have \( f(1) = \log(1+\sqrt{2}) \).

Step 6: Choose the Correct Answer

Compare \( f(1) = \log(1 + \sqrt{2}) \) with the available options. It matches (A) \( \log(1 + \sqrt{2}) \). Therefore, the correct answer is option (A).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus used to simplify complex integrals by changing variables. In this exercise, we encountered an integral that seemed quite complicated at first glance. To handle this, we employed substitution. By setting \( t = \sqrt{1 + x^2} \), we transformed the original variable \( x \) into a new variable \( t \). This process simplified the integrand significantly.

Here's why substitution is useful:

• It can transform an integrand into a simpler form that's easier to integrate.
• It often turns complex rational or trigonometric expressions into algebraic ones.
• The new variables often make partial fraction decomposition or other integration techniques easier.

After substitution, integrals in terms of \( t \) become more manageable, allowing us to proceed with partial fraction decomposition.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to integrate. In this problem, after substitution, we were left with an integral of \( \frac{t^2 - 1}{t(t+1)} \).

The steps for applying partial fraction decomposition involve:

• Expressing the fraction as a sum of simpler fractions: \( \frac{A}{t} + \frac{B}{t+1} \).
• Finding constants \( A \) and \( B \) using algebra to match coefficients, as shown in our step-by-step solution.

This technique allowed us to rewrite the integrand as \( \frac{-1}{t} + \frac{1}{t+1} \), which is much easier to handle when integrating. This is a standard approach when dealing with polynomials in the denominator that can be factored into linear terms.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, two of the main operations in calculus. It states that if you have a function \( f(x) \) that is continuous over an interval \([a, b]\), and \( F(x) \) is an antiderivative of \( f(x) \), then:\[ \int_a^b f(x) \, dx = F(b) - F(a) \]
In our scenario, once we found the antiderivative of the integrand using substitution and partial fraction decomposition, we applied the theorem to evaluate the definite integral from 0 to 1.

Here's how it works:

• Evaluate the antiderivative at the upper limit of the interval \( x = 1 \).
• Subtract the evaluation of the antiderivative at the lower limit \( x = 0 \).

This straightforward approach provides the difference between the values of the antiderivative at the boundaries of the interval, giving the final solution \( f(1) = \log(1 + \sqrt{2}) \).

Integration Techniques

Integration techniques are critical for evaluating integrals, especially those that aren't easily solvable at first glance. They include various methods, such as substitution, integration by parts, and partial fraction decomposition. In solving this exercise, we utilized substitution and partial fraction decomposition as primary techniques.

Here's why these techniques are essential:

• Substitution: Converts complex integrals into simpler forms by changing variables, as seen with \( t = \sqrt{1 + x^2} \).
• Partial Fraction Decomposition: Breaks down rational expressions into simpler components for easier integration.

Combining these tools, we can tackle a wide variety of integrals, leading to solutions that are not just accurate but also elegant and insightful. Mastery of these techniques ensures you can address both seemingly intractable and straightforward problems with ease.