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: Plan:
Think of the graph of $\ln(x^2+1)$.

Step 2: Steps:
The function $g(x) = \frac12\ln(1+x^2)$ is even with its minimum at $x = 0$. It falls for negative $x$ and rises for positive $x$, as $1+x^2$ shrinks towards $0$ and then grows.
So it is monotonically increasing on $(0,\infty)$, i.e. $R^+$.

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