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

Step 1: Start by simplifying the integral:

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

We recognize that the integrand suggests a standard substitution.

Step 2: Notice the form of the integrand:

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

This suggests that we should differentiate \(\log(x + \sqrt{1 + x^2})\) with respect to \(x\).

Step 3: Use the chain rule to compute the derivative of \(\log(x + \sqrt{1 + x^2})\). 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 \(1 + \frac{x}{\sqrt{1 + x^2}}\), and simplifying the expression yields:

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

Step 4: With this derivative, we can now directly integrate the original equation:

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

By recognizing the integral form, we conclude that:

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

Step 5: Since we are given that \(I = f(g(x)) + c\), and \(f(x) = \frac{x^2}{2}\) and \(g(x) = \log(x + \sqrt{1 + x^2})\), the correct answer is:

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

Answer 2

Evaluating the Integral $\int \frac{\log_e(x + \sqrt{1+x^2})}{\sqrt{1+x^2}} dx$

We are given the integral $\int \frac{\log_e(x + \sqrt{1+x^2})}{\sqrt{1+x^2}} dx = f(x) + c$. We need to determine the form of $f(x)$.

Step 1: Use Substitution

Let $t = \log_e(x + \sqrt{1+x^2})$.

Step 2: Find the differential $dt$

Differentiating $t$ with respect to $x$:

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

$$ \frac{dt}{dx} = \frac{1}{x + \sqrt{1+x^2}} \cdot (1 + \frac{x}{\sqrt{1+x^2}}) $$

$$ \frac{dt}{dx} = \frac{1}{x + \sqrt{1+x^2}} \cdot \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}} $$

$$ \frac{dt}{dx} = \frac{1}{\sqrt{1+x^2}} $$

So, $dt = \frac{dx}{\sqrt{1+x^2}}$.

Step 3: Substitute into the Integral

Substituting $t$ and $dt$ into the original integral:

$$ \int \frac{\log_e(x + \sqrt{1+x^2})}{\sqrt{1+x^2}} dx = \int t \, dt $$

Step 4: Evaluate the Integral in terms of $t$

$$ \int t \, dt = \frac{t^2}{2} + c $$

Step 5: Substitute back for $x$

Substitute $t = \log_e(x + \sqrt{1+x^2})$ back into the result:

$$ \frac{(\log_e(x + \sqrt{1+x^2}))^2}{2} + c $$

Step 6: Identify $f(x)$

We are given that the integral equals $f(x) + c$. Therefore:

$$ f(x) = \frac{(\log_e(x + \sqrt{1+x^2}))^2}{2} $$

Step 7: Compare with the Options

The options are given in a form that suggests $f(x) = \frac{(g(x))^2}{2}$. From our result, $g(x) = \log_e(x + \sqrt{1+x^2})$.

(A) $f(x) = \frac{x^2}{2}, g(x) = \log_e(x + \sqrt{1+x^2})$ - Incorrect $f(x)$.

(B) $f(x) = \log_e(x + \sqrt{1+x^2}), g(x) = \frac{x^2}{2}$ - Incorrect form of $f(x)$.

(C) $f(x) = x^2, g(x) = \log_e(x + \sqrt{1+x^2})$ - Incorrect form and $f(x)$.

(D) $f(x) = \log_e(x - \sqrt{1+x^2}), g(x) = x^2$ - Incorrect argument for logarithm.

Assuming there is a typo in option (A) for $f(x)$ and it was intended to relate to $g(x) = \log_e(x + \sqrt{1+x^2})$, then option (A) provides the correct $g(x)$.

Final Answer: (A)