## Table - rational functions

Fill in the following table (determine the intervals on which the solution is valid). Express the last function using integration by parts with a recurrence relation in terms of $$\int \frac{1}{(x^2 + 1)^k} \, dx$$.

$$f(x)$$ $$F(x)$$
$$\frac{1}{x - \alpha}$$
$$\frac{1}{(x - \alpha)^k}$$; $$k > 1$$
$$\frac{2x + p}{x^2 + px + q}$$
$$\frac{1}{x^2 + px + q}$$; $$q > \frac{p^2}4$$
$$\frac{2x + p}{(x^2 + px + q)^k}$$; $$k>1$$
$$\frac{1}{(x^2 + 1)^{k+1}}$$; $$k \geq 1$$
• #### Result

To simplify the notation, the solutions are written without the constant.

$$f(x)$$ $$F(x)$$
$$\frac{1}{x - \alpha}$$ $$\ln|x - \alpha|$$
$$\frac{1}{(x - \alpha)^k}$$; $$k > 1$$ $$\frac{-1}{(k-1)(x - \alpha)^{k-1}}$$
$$\frac{2x + p}{x^2 + px + q}$$ $$\ln|x^2 + px + q|$$
$$\frac{1}{x^2 + px + q}$$; $$q > \frac{p^2}4$$ $$\frac{1}{\sqrt t} \operatorname{arctan} ((x + p/2)/\sqrt t)$$, where $$t = q - p^2/4$$
$$\frac{2x + p}{(x^2 + px + q)^k}$$; $$k>1$$ $$\frac{-1}{(k-1)(x^2 + px + q)^{k-1}}$$
$$\frac{1}{(x^2 + 1)^{k+1}}$$; $$k \geq 1$$ $$\frac{1}{2k} \left(\frac{x}{(1+x^2)^k} + (2k-1) \int \frac{1}{(x^2 + 1)^k} \, dx \right)$$