Evaluate $ \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x} $. I need to solve the following integral:
$$
I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.
$$
Wolfram Alpha gives the answer as $ \dfrac{\pi}{2 \sqrt{2}} $.
I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?
 A: Let $u=\sqrt{x-1}$, $du=\frac{1}{2\sqrt{x-1}}$, then
$$I=2\int_0^\infty\frac{u^2}{(u^2+2)^2}du$$
We can apply partial fractions here. 
$$=2\int_0^\infty\left(\frac{1}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$
$$=2\int_0^\infty\frac{1}{u^2+2}du-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
The first integrand is almost $\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\frac{u}{\sqrt{2}}$.
$$I=\sqrt 2\int_0^\infty\frac{1}{s^2+1}ds-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
$$=\frac{\pi}{\sqrt 2}-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
Now to tackle the second integral we can use a trig sub. 
Let $u=\sqrt{2}\tan (t)$, $du=\sqrt 2 \sec^2(t)dt$.
$$I=\frac{\pi}{\sqrt 2}-4\sqrt 2 \int_0^{\pi/2}\frac{\sec ^2 t}{4\sec^4t}dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2\int _0^{\pi/2}\cos^2(t) dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2 \int_0^{\pi/2}\left(\frac{1}{2}\cos(2t)+\frac{1}{2} \right)dt$$ 
If we substitute $v=2t$ and split the integral up we get
$$I=\frac{\pi}{\sqrt 2}-\frac{1}{2\sqrt 2} \int_0^\pi \cos(v)dv-\frac{1}{\sqrt 2}\int_0^{\pi/2}1dt$$
The first integral clearly goes to 0 and the second integral becomes $\frac{\pi}{2\sqrt 2}$.
Therefore 
$$I=\frac{\pi}{2\sqrt 2}$$
A: Using a keyhole countour with the origin of the keyhole at $z=1$ and the small circle enclosing the value $z=1$ and using 
$$f(z) = \frac{\exp(1/2\log(z-1))}{(1+z)^2}$$
with the branch cut of the logarithm on the positive real axis and the argument from $0$ to $2\pi$ we get for the integral
$$I=\int_1^\infty \frac{\sqrt{x-1}}{(1+x)^2} dx$$ that
$$I(1-\exp(\pi i)) = 2\pi i\mathrm{Res}_{z=-1} f(z)$$
or
$$I = \pi i\mathrm{Res}_{z=-1} f(z).$$
Now the logarithmic term is certainly analytic in a neighborhood of $z=-1$
and we have $$\mathrm{Res}_{z=-1} f(z)
= \left.\left(\exp(1/2\log(z-1))\right)'\right|_{z=-1}
= \left.
\left(\exp(1/2\log(z-1))\right) \frac{1}{2}\frac{1}{z-1}\right|_{z=-1}
\\= \exp(1/2(\log 2 + \pi i)) \times -\frac{1}{2}\frac{1}{2}
= -\frac{1}{4} \sqrt{2} i.$$
This yields
$$I = -\frac{1}{4} \sqrt{2} i \times\pi i=\frac{\sqrt{2}\pi}{4}.$$
Remark. The estimates for the circular components are done using ML same as at this MSE link. We get for the large circle parameterized by $z=R e^{it}$
$$\lim_{R\rightarrow \infty} 2\pi R \frac{\sqrt{R+1}}{(R-1)^2} = 0.$$
The small circle is a parameterized with $z=1+\epsilon e^{it}$ and we get
$$\lim_{\epsilon\rightarrow 0} 2\pi\epsilon 
\frac{\sqrt{\epsilon}}{4} = 0.$$
A: $$I=\int_{1}^{\infty}\frac{\sqrt{x-1}}{(x+1)^2}dx$$  Use substitution $\frac{1}{x+1}=t$ which implies $\frac{dx}{(x+1)^2}=-dt$  So
$$I=-\int_{0.5}^{0}\sqrt{\frac{1}{t}-2}\:dt=\int_{0}^{0.5}\frac{\sqrt{1-2t}}{\sqrt{t}}dt$$
Again use substitution $\sqrt{t}=y$ which implies $\frac{dt}{\sqrt{t}}=2dy$
$$I=2\int_{0}^{\sqrt{0.5}}\sqrt{1-2y^2}dy=2\sqrt{2}\int_{0}^{\sqrt{0.5}}\sqrt{(\sqrt{0.5})^2-t^2}$$
Use standard integral $$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})$$
we get $$I=\frac{\pi}{2\sqrt{2}}$$
A: Hint Substitution $x=t^2+1$...
A: This is the general antiderivative. Just take the limits and you're good.
$$\int{\frac{\sqrt{x-1}}{(x+1)^2}dx}$$
Substitute $u = \sqrt{x-1}$
$$= \int{\frac{u}{(u^2+2)^2}du}$$
$$= 2\int\left(\frac{u}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$
$$= \int\frac{u}{\frac{u^2}{2}+1}du-4\int\frac{2}{(u^2+2)^2}du$$
Substitute $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}$
$$= \sqrt{2}\int\frac{1}{s^2+1}ds-4\int\frac{2}{(u^2+2)^2}du$$
$$= \sqrt{2}\arctan(s)-4\int\frac{2}{(u^2+2)^2}du$$
Substitute $u = \sqrt{2}\tan (p)\quad$ and $\quad du = \sqrt{2}\sec^2(p) dp\quad$ and $\quad(u^2 + 2)^2 = (2\tan^2(p)+2)^2 = 4\sec^4(p)\quad$ and $\quad p=\arctan\frac{u}{\sqrt{2}}$
$$= \sqrt{2}\arctan(s)-\sqrt{2}\int\cos^2(p)du$$
$$= \sqrt{2}\arctan(s)-\sqrt{2}\int\left(\frac{1}{2} \cos(2p) + \frac{1}{2}\right)du$$
$$= \sqrt{2}\arctan(s)-\frac{1}{\sqrt{2}}\int\cos(2p) - \frac{1}{\sqrt{2}}\int du$$
$$= \sqrt{2}\arctan(s)-\frac{p}{\sqrt{2}} - \frac{\sin(p)\cos(p)}{\sqrt{2}}$$
$$= \sqrt{2}\arctan(s)-\frac{\arctan\frac{u}{\sqrt{2}}}{\sqrt{2}} - \frac{\sin(\arctan\frac{u}{\sqrt{2}})\cos(\arctan\frac{u}{\sqrt{2}})}{\sqrt{2}}$$
Note that $\cos(\arctan(z)) = \frac{1}{\sqrt{z^2 + 1}}$ and $\sin(\arctan(z)) = \frac{z}{\sqrt{z^2+1}}$
$$= \frac{2\sqrt{2}(u^2+2)\arctan(s)+\sqrt{2}(u^2+2)\arctan\frac{u}{\sqrt{2}}+2u}{2(u^2+2)}$$
$$=\frac{\sqrt{2}(u^2+2)\arctan \frac{u}{\sqrt{2}}-2u}{2(u^2+2)}$$
$$=\frac{\sqrt{2}(x+1)\arctan \frac{\sqrt{x-1}}{\sqrt{2}}-2\sqrt{x-1}}{2(x+1)}$$
$$=\frac{\arctan \frac{\sqrt{x-1}}{\sqrt{2}}}{\sqrt{2}}-\frac{\sqrt{x-1}}{x+1}$$
A: I think I have a different answer. 
Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then
$$\begin{align}
\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx 
&= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\
&= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du
\end{align}$$
Now let $u = \frac{1}{t}$, so that $du = \frac{-1}{t^2} dt$. Continuing.
$$\begin{align}
\phantom{\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx} 
&= \frac{1}{\sqrt{2}} \int_0^1 t^2 (1-t)^\frac{1}{2} \frac{1}{\sqrt{t}} \frac{1}{t^2} dt \\
&= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{-1}{2}} (1-t)^\frac{1}{2} dt \\
&= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{2} - 1} dt \\
&= \frac{1}{\sqrt{2}} B ( \frac{1}{2} . \frac{3}{2} ) \\
&= \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{3}{2})} {\sqrt{2}\Gamma(2)} \\
&= \frac{\pi}{2\sqrt{2}}.
\end{align}$$
Original image of work by hand.
