Question

\(I=\int_1^2\dfrac{dx}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}\)

Answer 1

\(\dfrac{1}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\dfrac{\left(x+1\right)\sqrt{x}-x\sqrt{x+1}}{\left(x+1\right)^2x-x^2\left(x+1\right)}=\dfrac{\left(x+1\right)\sqrt{x}-x\sqrt{x+1}}{x\left(x+1\right)}\)

\(=\dfrac{\sqrt{x}}{x}-\dfrac{\sqrt{x+1}}{x+1}=x^{-\dfrac{1}{2}}-\left(x+1\right)^{-\dfrac{1}{2}}\)

Do đó:

\(I=\int\limits^2_1\left[x^{-\dfrac{1}{2}}-\left(x+1\right)^{-\dfrac{1}{2}}\right]dx=\left(2\sqrt{x}-2\sqrt{x+1}\right)|^2_1=...\)