One end of the diameter of the circle $x^2+y^2-6 x-5 y-1=0$ is $(-1,3)$, then the equation of the tangent at the other end of the diameter is

If $\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)$, then $x$ is equal to

The equation of the normal to the curve $y=x \log x$ parallel to $2 x-2 y+3=0$ is

$$\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x,$$ where $x>0$ is