Can someone help me why this is true:
$$\int_1^\infty \frac{1}{x(x+p)}\,dx = \frac{1}{p}\int_1^\infty\left(\frac{1}{x}-\frac{1}{x+p}\right)dx$$
2026-04-30 09:16:34.1777540594
Showing the integral $\int_1^\infty \frac{1}{x(x+p)}\,dx$ is convergent for $p$ greater than $-1$.
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Given that partial fraction decomposition, we have that a primitive for $$ \frac{1}{x(x+p)}=\frac{1}{p}\left(\frac{1}{x}-\frac{1}{x+p}\right)$$ is given by: $$ F(x)=\frac{1}{p}\,\log\left(\frac{x}{x+p}\right) $$ and such a function is bounded on $[1,+\infty)$ and converging to zero iff $p>-1$.