\(I=\int\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}=\int\sqrt{x+1}dx+\int\sqrt{x}dx\)
Xet \(I_1=\int\sqrt{x+1}dx\)
\(t=x+1\Rightarrow dt=dx\Rightarrow I_1=\int\sqrt{t}.dt=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+C\)
\(\Rightarrow I=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+\dfrac{2}{3}x^{\dfrac{3}{2}}+C\)
P/s: Bạn tự thay cận vô ạ