Question

If \(\int \frac{x+1}{x^2+1}dx = tan^{-1}x+g(x)+c\), where \(c\) is constant of integration, then the function \(g(x)\) is monotonically increasing in the interval

• A. \(R^+\)
• B. \(R^-\)
• C. \(R\)
• D. \(R-\{0\}\)

Hint:

Split the integrand; \(g(x)=\frac12\ln(x^2+1)\) and \(g'(x)=\frac{x}{x^2+1}\).

Answer 1

The Correct Option is A

Step 1: Split the integral:
\(\int\frac{x+1}{x^2+1}dx = \int\frac{x}{x^2+1}dx + \int\frac{1}{x^2+1}dx = \frac12\ln(x^2+1) + \tan^{-1}x + c\).
So \(g(x) = \frac12\ln(x^2+1)\).

Step 2: Monotonicity:
\(g'(x) = \frac{x}{x^2+1}\). The denominator is always positive, so the sign of \(g'\) is the sign of \(x\).
\(g'(x) > 0\) for \(x > 0\), i.e. \(g\) is increasing on \(R^+\). For \(x < 0\), \(g' < 0\), so \(g\) decreases on \(R^-\).

Step 3: Other options:
\(R^-\), \(R\) and \(R-\{0\}\) all include negative values, where \(g\) is decreasing. So only \(R^+\) works.

Final Answer:
The function \(g\) is increasing on \(R^+\), option (A). \[ \boxed{R^+} \]