tích phân

T

trantien.hocmai

$$I=\int_0^1 \dfrac{\sqrt{e^x}}{\sqrt{e^x+\dfrac{1}{e^x}}}dx \\
=\int_0^1 \dfrac{\sqrt{e^x}}{\dfrac{\sqrt{e^{2x}+1}}{\sqrt{e^x}}}dx \\
=\int_0^1 \dfrac{e^xdx}{\sqrt{e^{2x}+1}} \\ $$
$\text{đặt }u=e^x \rightarrow du=e^xdx \\
\text{đổi cận } x=0 \rightarrow u=1, x=1 \rightarrow u=e \\$
$$I=\int_1^e \dfrac{du}{\sqrt{u^2+1}}$$
$\text{đặt }t=u+\sqrt{u^2+1} \rightarrow dt=(1+\dfrac{u}{\sqrt{u^2+1}})du \\
=\dfrac{u+\sqrt{u^2+1}}{\sqrt{u^2+1}}du=\dfrac{tdu}{\sqrt{u^2+1}} \\
\rightarrow \dfrac{dt}{t}=\dfrac{du}{\sqrt{u^2+1}} \\
\text{tới đây thì quá dễ rồi}$
 
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