Chủ đề này có 3 trả lời

[thaitronganh1992]

$\int_{0}^{\sqrt{3}}\dfrac{1}{\sqrt{x{}^{2}+1}}dx$

[Crystal]

$\int_{0}^{\sqrt{3}}\dfrac{1}{\sqrt{x{}^{2}+1}}dx$

Gợi ý: đặt $x = tgt$

[thaitronganh1992]

Gợi ý: đặt $x = tgt$

em làm rồi mà hok ra a giải dùm em đi

[Crystal]

$\int_{0}^{\sqrt{3}}\dfrac{1}{\sqrt{x{}^{2}+1}}dx$

Đặt $x = tgt \Rightarrow dx = \dfrac{{dt}}{{{{\cos }^2}t}},\,\,x \in \left[ {0,\sqrt 3 } \right] \Rightarrow t \in \left[ {0,\dfrac{\pi }{3}} \right]$

Khi $x = 0 \Rightarrow t = 0;x = \sqrt 3 \Rightarrow t = \dfrac{\pi }{3}$
Khi đó: $$\int\limits_0^{\sqrt 3 } {\dfrac{1}{{\sqrt {{x^2} + 1} }}} dx = \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{1}{{\sqrt {t{g^2}t + 1} }}} \dfrac{{dt}}{{{{\cos }^2}t}} = \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{{\cos t}}{{{{\cos }^2}t}}dt = } \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{{d\left( {\sin t} \right)}}{{\left( {1 - {{\sin }^2}t} \right)}}} $$
$$ = \left. {\dfrac{1}{2}\ln \left| {\dfrac{{1 + \sin t}}{{1 - \sin t}}} \right|} \right|_0^{\dfrac{\pi }{3}} = \dfrac{1}{2}\left( {\ln \dfrac{{1 + \dfrac{{\sqrt 3 }}{2}}}{{1 - \dfrac{{\sqrt 3 }}{2}}}} \right) = \dfrac{1}{2}\ln \dfrac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} = \ln \left( {2 + \sqrt 3 } \right)$$
Vậy $$\int\limits_0^{\sqrt 3 } {\dfrac{1}{{\sqrt {{x^2} + 1} }}} dx = \boxed{\ln \left( {2 + \sqrt 3 } \right)}$$