Question

If \(\int e^{x+tan^{-1}x}(\frac{x^2+2}{sec^2(tan^{-1}x)})dx = e^{f(x)}+c\), then \(\ldots\)

• A. \(f(x)\) is strictly decreasing on \(R\).
• B. \(f(x)\) is strictly increasing on \(R^+\) and strictly decreasing on \(R^-\).
• C. \(f(x)\) is strictly increasing on \(R\).
• D. \(f(x)\) is strictly decreasing on \(R^+\) and strictly increasing on \(R^-\).

Hint:

Notice that the integrand is the derivative of e^(x + tan^-1 x). So f(x) = x + tan^-1 x.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need the function \(f(x)\) in the exponent of the result. If the integrand is the derivative of \(e^{f(x)}\), then \(f\) is read off directly.

Step 2: Key Formula or Approach:
\(\sec^2(\tan^{-1}x) = 1 + \tan^2(\tan^{-1}x) = 1 + x^2\). Also \(\dfrac{d}{dx}\left(x + \tan^{-1}x\right) = 1 + \dfrac{1}{1+x^2} = \dfrac{x^2+2}{1+x^2}\).

Step 3: Detailed Explanation:
The integrand is
\[ e^{x+\tan^{-1}x}\cdot\frac{x^2+2}{1+x^2} = e^{x+\tan^{-1}x}\cdot\frac{d}{dx}\left(x + \tan^{-1}x\right) \]
By the rule \(\int e^{g(x)}g'(x)\,dx = e^{g(x)} + c\), the integral is \(e^{x+\tan^{-1}x} + c\). So
\[ f(x) = x + \tan^{-1}x \]
Its derivative is \(f'(x) = \dfrac{x^2+2}{1+x^2}\), which is positive for every real \(x\), since both numerator and denominator are positive.
So \(f\) is strictly increasing on all of \(\mathbb{R}\). Options (A), (B) and (D) all claim decreasing behaviour somewhere, which contradicts \(f' > 0\).

Final Answer:
\(f(x)\) is strictly increasing on \(R\), option (C). \[ \boxed{\text{strictly increasing on } R \text{ (C)}} \]