Problem 138 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of the integral \(\int_{0}^{\infty} \frac{d x}{\left(x+\sqrt{x^{2}+1}\right)^{n}}\), where \(n>1\), is (A) \(\frac{n^{2}}{n^{2}-1}\) (B) \(\frac{n}{n^{2}-1}\) (C) \(\frac{n^{2}}{n^{2}+1}\) (D) \(\frac{n}{n^{2}+1}\)

Step-by-Step Solution

Answer

The value of the integral is (B) \( \frac{n}{n^2 - 1} \).

Step 1: Perform a substitution

First, substitute \( x = \sinh(t) \). Then, \( dx = \cosh(t) \ dt \) and \( x^2 + 1 = \cosh^2(t) \). Therefore, \( x + \sqrt{x^2 + 1} = \sinh(t) + \cosh(t) = e^t \). This transforms the integral to \( \int_{0}^{\infty} \frac{\cosh(t) \, dt}{(e^t)^n} \).

Step 2: Simplify the integral

The expression becomes \( \int_{0}^{\infty} e^{-nt} \cosh(t) \, dt \). Since \( \cosh(t) = \frac{e^t + e^{-t}}{2} \), we can write the integral as \( \frac{1}{2}\int_{0}^{\infty} (e^{-(n-1)t} + e^{-(n+1)t}) \, dt \).

Step 3: Evaluate the integral

We now evaluate each part separately. The integral \( \int_{0}^{\infty} e^{-(n-1)t} \, dt \) equals \( \frac{1}{n-1} \) and the integral \( \int_{0}^{\infty} e^{-(n+1)t} \, dt \) equals \( \frac{1}{n+1} \).

Step 4: Combine the results and simplify

Combining, we get \( \frac{1}{2} \left( \frac{1}{n-1} + \frac{1}{n+1} \right) = \frac{1}{2} \left( \frac{n+1 + n-1}{(n-1)(n+1)} \right) = \frac{n}{(n-1)(n+1)} \). This simplifies to \( \frac{n}{n^2 - 1} \).

Step 5: Match with options

The final simplified form \( \frac{n}{n^2 - 1} \) corresponds to option (B).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas instead of circles. They include functions like hyperbolic sine (\( \sinh(x) \)) and hyperbolic cosine (\( \cosh(x) \)). These are particularly useful in calculus due to their smooth exponential characteristics.
Hyperbolic sine and cosine are defined as:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

In this problem, we perform a substitution where \( x = \sinh(t) \) to transform our integral. This substitution simplifies the process by leveraging properties of \( \sinh(t) \) and \( \cosh(t) \) for calculation. It is notable that the relationship \( \cosh^2(t) = \sinh^2(t) + 1 \) is similar to \( \cos^2(x) = \sin^2(x) + 1 \) for trigonometric functions.
Ultimately, understanding hyperbolic functions can greatly simplify complex calculus problems, especially those involving exponential or growth-related components.

Integral Substitution

Integral substitution is a key technique to solve complex integrals. It involves changing variables to simplify the integral, making it easier to evaluate.
In this problem, the substitution \( x = \sinh(t) \) transforms both the variable of integration and the limits of the integral. Substitution results in the expression \( x + \sqrt{x^2 + 1} = e^t \), which allows us to rewrite the integral.
This technique requires:

• Identifying a suitable substitution.
• Expressing all parts of the integrand in terms of the new variable.
• Changing the limits of integration.

Through these steps, we can convert an initially intimidating integral into one that is manageable and straightforward to solve. By using \( dt = \frac{dx}{\cosh(t)} \), we tailor the integral to a form suitable for direct evaluation. Integral substitutions are powerful tools for handling challenging calculus problems.

Infinite Interval Integration

Infinite interval integration involves evaluating integrals with either one or both limits being infinite. Such integrals often appear in contexts involving limits approaching infinity.
To solve an integral across an infinite range like \( \int_{0}^{\infty} \), as seen in this exercise, often involves finding the antiderivative and then evaluating the limit as the interval extends to infinity.
Key strategies include:

• Examining the behavior of the integrand as it approaches infinity.
• Using convergence tests and properties of exponential decay, ensuring the integral exists.
• Calculating using properties such as \( \int_{0}^{\infty} e^{-at} \, dt = \frac{1}{a} \) for \( a > 0 \).

Such techniques are critical, as they verify the convergence and allow us to handle situations where conventional methods fall short. By addressing integrals over infinite intervals, we extend the reach of calculus into areas like probability, physics, and beyond.