I'm trying to evaluate:
$$\lim_{c \rightarrow \infty} \int_{-c}^c \frac{1+x}{1+x^2}dx$$
but I don't understand how to evaluate
$$\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$$
How?
2026-05-06 09:40:12.1778060412
How to evaluate $\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$
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Hint. Observe that $$ \int_{-c}^c \frac{1+x}{1+x^2}dx=\int_{-c}^c \frac{1}{1+x^2}dx=2\int_0^c \frac{1}{1+x^2}dx=2\arctan(c). $$ Can you take it from here?
Some details. We have $$ \begin{align} \color{blue}{\int_{-c}^c \frac{x}{1+x^2}dx}&=\int_{-c}^0 \frac{x}{1+x^2}dx+\int_0^c \frac{x}{1+x^2}dx \\\\&=\int_{c}^0 \frac{-x}{1+(-x)^2}(-dx)+\int_0^c \frac{x}{1+x^2}dx \\\\&=-\int_0^c \frac{x}{1+x^2}dx+\int_0^c \frac{x}{1+x^2}dx \\\\&=\color{blue}{0} \end{align} $$ giving $$ \int_{-c}^c \frac{1+x}{1+x^2}dx=\int_{-c}^c \frac{1}{1+x^2}dx+\color{blue}{\int_{-c}^c \frac{x}{1+x^2}dx}=\int_{-c}^c \frac{1}{1+x^2}dx+\color{blue}{0}. $$