Question

If \(\int \frac{\log(x + \sqrt{1 + x^2})}{1 + x^2} \, dx = f(g(x)) + c\), then:

• A. \(f(x) = \frac{x^2}{2}, \, g(x) = \log(x + \sqrt{1 + x^2})\)
• B. \(f(x) = \log(x + \sqrt{1 + x^2}), \, g(x) = \frac{x^2}{2}\)
• C. \(f(x) = x^2, \, g(x) = \log(x + \sqrt{1 + x^2})\)
• D. \(f(x) = \log(x - \sqrt{1 + x^2}), \, g(x) = x^2\)

Hint:

When dealing with integrals of logarithmic functions, look for substitutions involving the argument of the logarithm and its derivative. This can simplify the problem significantly

Answer 1

The Correct Option is A

Begin by simplifying the integral:

\( I = \int \frac{\log(x + \sqrt{1 + x^2})}{1 + x^2} \, dx \)

The integrand's structure implies a common substitution method.

Observe the integrand's form:

\(\frac{d}{dx} \left(\log(x + \sqrt{1 + x^2})\right) = \frac{1}{1+x^2}\).

This suggests differentiating \(\log(x + \sqrt{1 + x^2})\) with respect to \(x\).

Calculate the derivative of \(\log(x + \sqrt{1 + x^2})\) using the chain rule: The derivative is:

\(\frac{d}{dx} \log(x + \sqrt{1 + x^2}) = \frac{1}{x + \sqrt{1 + x^2}} \cdot \frac{d}{dx}(x + \sqrt{1 + x^2})\).

The derivative of \(x + \sqrt{1 + x^2}\) is \(1 + \frac{x}{\sqrt{1 + x^2}}\), leading to:

\(\frac{d}{dx} \log(x + \sqrt{1 + x^2}) = \frac{1}{1 + x^2}\).

Using this derivative, directly integrate the original equation:

\( I = \int \log(x + \sqrt{1 + x^2}) \cdot \frac{1}{1 + x^2} \, dx \).

The integral's form leads to:

\( I = \frac{x^2}{2} + c \).

Given \(I = f(g(x)) + c\), and \(f(x) = \frac{x^2}{2}\) and \(g(x) = \log(x + \sqrt{1 + x^2})\), the answer is:

\( f(x) = \frac{x^2}{2}, \, g(x) = \log(x + \sqrt{1 + x^2}) \).