1. Ta có:
$\int\limits f(x)dx=\int\limits (\tan x+ \frac{1}{\sqrt{2x+1}+\sqrt{2x-1} } )dx$
$=\int\limits \frac{\sin x.dx}{\cos x} + \int\limits \frac{\sqrt{2x+1}-\sqrt{2x-1} }{2} dx$
$=-\ln|\cos x|+\frac{1}{3}[(2x+1)^{\frac{3}{2}}-(2x-1)^{\frac{3}{2} } +C $
2. Sử dụng đồng nhất thức $x=x+1-1$, ta được:
$f(x)=\frac{x+1-1}{\sqrt[10]{x+1} } =(x+1)^{ \frac{9}{10}}-(x+1)^{-\frac{1}{10} } $
Do đó:
$\int\limits f(x)dx=\int\limits [(x+1)^{ \frac{9}{10}}-(x+1)^{-\frac{1}{10} } ]dx=\frac{10}{19}(x+1)^{ \frac{19}{10}}-\frac{10}{9}(x+1)^{ \frac{9}{10}}+C $