Đáp án:
$\dfrac12\left(\sqrt x +\sqrt{x+1}\right) - \dfrac12\ln\left(\sqrt x +\sqrt{x+1}\right)- \dfrac{1}{2\left(\sqrt x +\sqrt{x+1}\right)} + \dfrac{1}{4\left(\sqrt x +\sqrt{x+1}\right)^2}+ C$
Giải thích các bước giải:
$\quad I =\displaystyle\int\dfrac{1}{\sqrt{x+1} +\sqrt x +1}dx$
Đặt $t = \sqrt x + \sqrt{x+1}$
$\to \dfrac{1}{t}= \sqrt{x+1} -\sqrt x$
$\to t -\dfrac1t = 2\sqrt x$
$\to \dfrac14\left(t-\dfrac{1}{t}\right)^2 = x$
$\to \dfrac12\left(1 +\dfrac{1}{t^2}\right)\left(t -\dfrac1t\right)dt = dx$
$\to \dfrac{t^4 -1}{2t^3}dt = dx$
Ta được:
$\quad I =\displaystyle\int\dfrac{1}{1+t}\cdot \dfrac{t^4 -1}{2t^3}dt$
$\to I = \dfrac12\displaystyle\int\left(1 -\dfrac1t +\dfrac{1}{t^2} -\dfrac{1}{t^3}\right)dt$
$\to I = \dfrac12t-\dfrac{1}{2}\ln|t| - \dfrac{1}{2t} +\dfrac{1}{4t^2} + C$
$\to I = \dfrac12\left(\sqrt x +\sqrt{x+1}\right) - \dfrac12\ln\left(\sqrt x +\sqrt{x+1}\right)- \dfrac{1}{2\left(\sqrt x +\sqrt{x+1}\right)} +\dfrac{1}{4\left(\sqrt x +\sqrt{x+1}\right)^2}+ C$
