<s>[INST] 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.
 [/INST] 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$$
I= -\int_0^{\frac{1}{\sqrt 2}  } \frac{d c}{(1+c) c^2}= \int_0^{\frac{1}{\sqrt{2}}}\left(\frac{1}{c}-\frac{1}{c^2}-\frac{1}{1+c}\right) d c= \sqrt{2}-1+\ln [2(\sqrt{2}-1)]
$$
</s>