If int frac{x + 1}{x^2 + 1} dx = tan^{-1} x + | vedclass

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(A) Given $\int \frac{x + 1}{x^2 + 1} dx = \int \frac{x}{x^2 + 1} dx + \int \frac{1}{x^2 + 1} dx$.$= \frac{1}{2} \ln(x^2 + 1) + 	an^{-1} x + c$.Compa

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$R^+$

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(A) Given $\int \frac{x + 1}{x^2 + 1} dx = \int \frac{x}{x^2 + 1} dx + \int \frac{1}{x^2 + 1} dx$.$= \frac{1}{2} \ln(x^2 + 1) + 	an^{-1} x + c$.Comparing this with the given expression $	an^{-1} x + g(x) + c$, we get $g(x) = \frac{1}{2} \ln(x^2 + 1)$.To find where $g(x)$ is monotonically increasing, we find its derivative: $g'(x) = \frac{1}{2} \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{x}{x^2 + 1}$.For $g(x)$ to be monotonically increasing, $g'(x) > 0$.$\frac{x}{x^2 + 1} > 0$. Since $x^2 + 1 > 0$ for all $x \in R$, we must have $x > 0$.Thus, $g(x)$ is monotonically increasing in the interval $(0, \infty)$, which is $R^+$.

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

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