## Calculus 8th Edition

$\Sigma \frac{n}{n^2+1}$ $\int_{1}^{\infty}\frac{x}{x^2+1}dx$ $u=x^2+1$ $du=2xdx$ gives $\frac{1}{2x}du=dx$ $\int_{1}^{\infty}\frac{x}{x^2+1}dx=\int_{1}^{\infty}\frac{x}{2xu}du=\int_{2}^{\infty}\frac{1}{u}du=[\frac{1}{2}ln(u)]_{2}^{\infty}$ $\lim\limits_{t \to \infty}[\frac{1}{2}ln(u)]_{2}^{t}=\infty-\frac{1}{2}ln(2)=\infty$ Divergent