Find the value of the integral: $\int \frac{2x^2}{x^3+1} dx$

If $f(x)+k$ is obtained by evaluating $\int \frac{x^3}{\left(1+x^2\right)^3} d x$ using the substitution $x=\tan \theta$, and $g(x)+c$ is obtained by evaluating $\int \frac{x^3}{\left(1+x^2\right)^3} d x$ using the substitution $x^2+1=z$, then $f(x)-g(x)+k-c=$

$\int \frac{1}{(\cos^{-1} x) \sqrt{1 - x^2}} dx = $

$\int e^{\sqrt{x}} \, dx = $ . . . . . . $+ c ; x > 0$

$\int \frac{\cos x-\sin x}{8-\sin 2x} dx = \frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c$,then $p = \ldots$

$\int {{x^3}\sqrt {3 + 5{x^4}} } \;dx = $