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Question

$\int\limits_{\sqrt{3}}^{\sqrt{8}}\frac{dx}{1+x+\sqrt{1+x^2}}$

shinigem_1994 · Answer 1 · 5/20/2013 10:11:47 AM

$\frac{1}{1+x+\sqrt{1+x^2}}=\frac{1+x-\sqrt{1+x^2}}{2x}=\frac{1}{2x}+\frac{1}{2}-\frac{x\sqrt{1+x^2}}{2x^2}$

$\Rightarrow I=\int\limits_{\sqrt{3}}^{\sqrt{8}}(\frac{1}{2x}+\frac{1}{2}-\frac{x\sqrt{1+x^2}}{2x^2})dx=(\frac{1}{2}ln|x|+\frac{1}{2}x)\left| \begin{matrix} \sqrt{8}\\ \sqrt{3} \end{matrix}{} \right.-\int\limits_{a}^{b}\frac{x\sqrt{1+x^2}}{2x^2}$

$=\frac{1}{4}ln\frac{8}{3}+\frac{1}{2}(\sqrt{8}-\sqrt{3})-\int\limits_{\sqrt{3}}^{\sqrt{8}}\frac{x\sqrt{1+x^2}}{2x^2}dx$

đặt

$t=\sqrt{1+x^2}\Rightarrow 2tdt=2xdx\Rightarrow tdt=xdt$

đổi cận

$x=\sqrt{3}\Rightarrow t=2,x=\sqrt{8}\Rightarrow t=3$

$\Rightarrow I=\frac{1}{4}ln\frac{8}{3}+\frac{1}{2}(\sqrt{8}-\sqrt{3})-\int\limits_{2}^{3}\frac{t^2dt}{2(t^2-1)}$

ta có

$\int\limits_{2}^{3}\frac{t^2}{t^2-1}=\int\limits_{2}^{3}(1+\frac{1}{2(t-1)}-\frac{1}{2(t+1)})dt=t+\frac{1}{2}ln|\frac{t-1}{t+1}|=1+\frac{1}{2}ln\frac{3}{2}$

$I=\frac{1}{2}ln\frac{4}{3}+\frac{\sqrt{8}-\sqrt{3}-1}{2}$

binhnguyenhoangvu · Answer 2 · 5/20/2013 10:12:00 AM

$I = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{dx}}{{1 + x + \sqrt {1 + {x^2}} }}}$

Nhân lượng liên hợp, ta được:

$I = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{1 + x - \sqrt {1 + {x^2}} }}{{2x}}} dx = \frac{1}{2}\int\limits_{\sqrt 3 }^{\sqrt 8 } {\left( {1 + \frac{{1 - \sqrt {1 + {x^2}} }}{x}} \right)} dx$

$= \frac{1}{2}(\sqrt 8 - \sqrt 3 ) - \frac{1}{2}\int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} dx$

Đặt

$K = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} dx$

Lại nhân lượng liên hợp, ta được:

$K = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{x}{{\sqrt {1 + {x^2}} + 1}}} dx = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{\sqrt {1 + {x^2}} }}{{(\sqrt {1 + {x^2}} + 1)}}.\frac{x}{{\sqrt {1 + {x^2}} }}} dx$

Đặt