converge/diverge: int[1,infty][1/((x+1)^2 sqrt[x-1])dx

[cheffy]

\(\displaystyle \[ \int_1^\infty \frac{1}{(x+1)^2 \sqrt{x-1} } \,dx.\]\)

Does this converge or diverge? I don't know how to rationalize this.

Thanks.

 

[pka]

This integral certainly converges. What do you need to know?

\(\displaystyle \L \begin{array}{l}
1 < x < 2\quad \Rightarrow \quad 4 < \left( {x + 1} \right)^2 < 9\quad \\
\Rightarrow \quad \int\limits_1^2 {\frac{{dx}}{{\left( {x + 1} \right)^2 \sqrt {x - 1} }}} > \int\limits_1^2 {\frac{{dx}}{{4\sqrt {x - 1} }}} \\
\\
x > 2\quad \Rightarrow \quad \sqrt {x - 1} > 1 \\
\quad \Rightarrow \quad \int\limits_2^\infty {\frac{{dx}}{{\left( {x + 1} \right)^2 \sqrt {x - 1} }}} < \int\limits_2^\infty {\frac{{dx}}{{\left( {x + 1} \right)^2 }}} \\
\end{array}\)

 

[cheffy]

So because \(\displaystyle \[ \int_1^2 \frac{1}{4 \sqrt{x-1} \,dx.\]\) converges then \(\displaystyle \[ \int_1^2 \frac{1}{(x+1)^2 \sqrt{x-1} \,dx.\]\) converges too? And the same for the other pair?

 

[pka]

Exactly.