Integrate solution: \int(\frac{\tan^{-1}x}{x-\tan^{-1}x})^2dx

How you would find the equation of a line normal to a curve?

Compute the gradient of the following functions and evaluate it at the given point P.
${\displaystyle g(x,y)=x^2-4x^{2}y-8x{y}^2;P(-1,2)}$

How do you determine whether the function ${\displaystyle f\left(x\right)=\frac{2x-3}{x^2}}$ is concave up or concave down and its intervals?

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
${\displaystyle h\left(t\right)=\frac{2t-3t^2}{1+t^3},a=1}$

How do you find the concavity for ${\displaystyle f\left(x\right)=-6\sqrt{x}}$?

Finding an antiderivative
I need to find the following:
$\int \sqrt{\frac{1-x}{1+x}}\cdot \frac{1}{x}\mathrm{d}x$
Firstly, it has to be that $x\in (-1,0)\cup (0,1]$. From this, it is implied that $1+x>0$. Also, it holds that $1-x\ge 0$.
What I did is that I multiplied both numerator and denominator by $\sqrt{1+x}$. Thus, we have:
$\int \frac{\sqrt{1-x^2}}{1+x}\cdot \frac{1}{x}\mathrm{d}x\begin{array}{l}{\displaystyle \stackrel{x=\sin t}{=}\int \frac{\cos t}{1+\sin t}\cdot \frac{1}{\sin t}\cdot \cos t\mathrm{d}t}\\ ={\displaystyle \int \frac{{\cos }^{2}t}{(1+\sin t)\sin t}\mathrm{d}t}\\ ={\displaystyle \int \frac{1-{\sin }^{2}t}{(1+\sin t)\sin t}\mathrm{d}t}\\ ={\displaystyle \int \frac{(1-\sin t)(1+\sin t)}{(1+\sin t)\sin t}\mathrm{d}t={\displaystyle \int \frac{1}{\sin t}-1\mathrm{d}t.}}\end{array}$
Thus, the final answer is:
$\log |\tan \left(\frac{\arcsin x}{2}\right)|-\arcsin x+C,$,
since $\int \frac{1}{\sin t}\mathrm{d}t=\log |\tan \left(\frac{t}{2}\right)|+c.$.
The answer given in the textbook is:
$\log \left|\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}\right|+2\arctan \left(\sqrt{\frac{1-x}{1+x}}\right)+constant$
I have derived that the arguments in the logarithms are the same, but it seems difficult to connect arc sin x with $2\arctan \left(\sqrt{\frac{1-x}{1+x}}\right).$.

Concave or convex at ${\displaystyle x=-1}$?
${\displaystyle f\left(x\right)=x^4-4x^3+x-4}$