Tính nguyên hàm `∫1/(sqrt(x+1)+sqrtx+1)dx`

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:

$\quad I =\displaystyle\int\dfrac{1}{1+t}\cdot \dfrac{t^4 -1}{2t^3}dt$

$\to I = \dfrac12\displaystyle\int\left(1 -\dfrac1t +\dfrac{1}{t^2} -\dfrac{1}{t^3}\right)dt$

$\to I = \dfrac12t-\dfrac{1}{2}\ln|t|  - \dfrac{1}{2t} +\dfrac{1}{4t^2} + C$

$\to I = \dfrac12\left(\sqrt x +\sqrt{x+1}\right) - \dfrac12\ln\left(\sqrt x +\sqrt{x+1}\right)- \dfrac{1}{2\left(\sqrt x +\sqrt{x+1}\right)} +\dfrac{1}{4\left(\sqrt x +\sqrt{x+1}\right)^2}+ C$

Đáp án:

Giải thích các bước giải: Một cách để tham khảo

$\dfrac{1}{\sqrt{x + 1} + \sqrt{x} + 1} = \dfrac{(\sqrt{x} + 1) - \sqrt{x + 1}}{(\sqrt{x} + 1)² - (\sqrt{x + 1})²}$

$ = \dfrac{\sqrt{x} + 1 - \sqrt{x + 1}}{2\sqrt{x}} = \dfrac{1}{2} + \dfrac{1}{2\sqrt{x}} - \dfrac{\sqrt{x + 1}}{2\sqrt{x}}$

Đặt $: t = \sqrt{x} ⇒ t² = x; dt = \dfrac{dx}{2\sqrt{x}}$

Tính $:∫\dfrac{\sqrt{x + 1}}{2\sqrt{x}}dx = ∫\sqrt{t² + 1}dt$

$ = \dfrac{1}{2}t\sqrt{t² + 1} + \dfrac{1}{2}ln(t + \sqrt{t² + 1}) + C_{1}$

$ = \dfrac{1}{2}\sqrt{x(x + 1)} + \dfrac{1}{2}ln(\sqrt{x} + \sqrt{x + 1}) + C_{1} $

$ ⇒ ∫\dfrac{1}{\sqrt{x + 1} + \sqrt{x} + 1}dx = ∫\dfrac{1}{2}dx + ∫\dfrac{1}{2\sqrt{x}}dx - ∫\dfrac{\sqrt{x + 1}}{2\sqrt{x}}dx$

$ = \dfrac{x}{2} + \sqrt{x} - \dfrac{1}{2}\sqrt{x(x + 1)} - \dfrac{1}{2}ln( \sqrt{x} + \sqrt{x + 1}) + C$