$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \, dx = $

For real numbers $\alpha, \beta, \gamma$ and $\delta,$ if $\int \frac{\left(x^{2}-1\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}\left(\frac{x^{2}+1}{x}\right)} d x =\alpha \log _{e}\left(\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right) +\beta \tan ^{-1}\left(\frac{\gamma\left(x^{2}-1\right)}{x}\right)+\delta \tan ^{-1}\left(\frac{x^{2}+1}{x}\right)+C$ where $C$ is an arbitrary constant,then the value of $10(\alpha+\beta \gamma+\delta)$ is equal to ....... .

If $\int \sec ^2 x \operatorname{cosec}^4 x \, dx = -\frac{1}{3} \cot ^3 x + k \tan x - 2 \cot x + C$,then $k$ is equal to

Let $I(x)=\int\frac{3dx}{(4x+6)(\sqrt{4x^{2}+8x+3})}$ and $I(0)=\frac{\sqrt{3}}{4}+20$. If $I(\frac{1}{2})=\frac{a\sqrt{2}}{b}+c$, where $a, b, c \in N$ and $gcd(a,b)=1$, then $a+b+c$ is equal to:

$\int \frac{x}{\sqrt{x^2-2x+5}} dx=$

If $\int \frac{d x}{\left(x^{2}+x+1\right)^{2}}=a \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+b\left(\frac{2 x+1}{x^{2}+x+1}\right)+C$ for $x>0$,where $C$ is the constant of integration,then the value of $9(\sqrt{3} a+b)$ is equal to ... .