Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\sec \psi}{\tan \psi+1+\sec \psi}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi - 1}{\tan \psi+\sec \psi+1}d\psi$
Now Multiply both Numerator and Denominator by $\left(\tan \psi+\sec \psi-1\right)$
So $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi+1} \times \frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi-1}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\left(\tan \psi+\sec \psi-1\right)^2}{\left(\tan \psi+\sec \psi\right)^2-1}d\psi$
Now I did not understand how can i solve after that
Help required
Thanks.
 A: Hint
As suggested by Fabian, multiply both numerator and denominator by $1+x -\sqrt{1+x^2}$. After simplfication, you should arrive to $$\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}-1}{x}$$
I am sure that you can take from here.
A: Solution:
Let $$t=x+1+\sqrt{1+x^2}\to 1+x^2=(t-1-x)^2\to x=\frac{t^2-2t}{2t-2}\to dx=\frac{t^2-2t+2}{2(t-1)^2}dt$$
We have $$x-1+\sqrt{1+x^2}=t-2$$
And $$\frac{x-1+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}=\frac{t-2}{t}$$
Hence $$I=\int_{0}^{1}\frac{x-1+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}dx=\int_{2}^{2+\sqrt{2}}\frac{t-2}{t}\frac{t^2-2t+2}{2(t-1)^2}dt$$
$$=\int_{2}^{2+\sqrt{2}}\left[\frac{1}{2}-\frac{2}{t}+\frac{1}{t-1}-\frac{1}{2(t-1)^2}\right]dt=\cdots$$
A: Setting $x=\tan2y$  $$F=\frac{\tan2y+\sec2y-1}{\tan2y+\sec2y+1}=\frac{\sin2y+1-\cos2y}{\sin2y+1+\cos2y}$$
Using Double Angle formula,  $$F=\frac{2\sin y\cos y+2\sin^2y}{2\sin y\cos y+2\cos^2y}=\frac{2\sin y(\cos y+\sin y)}{2\cos y(\sin y+\cos y)}=\tan y$$ assuming  $\cos y+\sin y\ne0\iff\tan y\ne-1\implies x=\tan2y\ne\infty$
A: After multiplying the integrand with $1+x-\sqrt{1+x^2}$, then you may use Euler substitution to evaluate the integral.
For further learning, you may refer to this link. The integral in the given link almost similar with yours, perhaps yours is simpler.
A: Let $x=\sqrt{y^2-1}$ to reduce the integral
\begin{align}
&\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx
=\int_1^{\sqrt2}\frac y{1+y}dy=\sqrt2-1+\ln[2(\sqrt2-1)]
\end{align}
A: Letting $x=\tan \theta$ gives
$$
\begin{aligned}
\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}} &=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1+\sec \theta} \\
&=\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta+1} \\
&=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+2 \cos ^2 \frac{\theta}{2}} \\
&=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2} }=\frac{\sin \theta}{1+\cos \theta}
\end{aligned}
$$
Then let $c=\cos \theta,$ we have$$
I= -\int_0^{\frac{1}{\sqrt 2}  } \frac{d c}{(1+c) c^2}= \int_0^{\frac{1}{\sqrt{2}}}\left(\frac{1}{c}-\frac{1}{c^2}-\frac{1}{1+c}\right) d c= \sqrt{2}-1+\ln [2(\sqrt{2}-1)]
$$
