cận 1----->2 \int_{}^{}(x^2+2)/x *căn(x^4-x^2+4)..............
[TEX]I=\int_{1}^{2}\frac{x^2+2}{x.\sqrt{x^4-x^2+4}}dx=\int_{1}^{2}\frac{x^2}{x.\sqrt{x^4-x^2+4}}dx+\int_{1}^{2}\frac{2}{x.\sqrt{x^4-x^2+4}}dx = E+2F[/TEX]
[TEX]E=\int_{1}^{2}\frac{x^2}{x.\sqrt{x^4-x^2+4}}dx=\int \frac{xdx}{\sqrt{(x^2-1)^2+(x^2-1)+4}}[/TEX]
[TEX]Dat \ x^2-1=t \Rightarrow 2xdx=dt [/TEX]
[TEX]\Leftrightarrow E=\frac{1}{2}\int \frac{dt}{\sqrt{t^2+t+4}}[/TEX]
[TEX]F=\int_{1}^{2}\frac{1}{x.\sqrt{x^4-x^2+4}}dx \\\\ Dat \ x=\frac{1}{t}\Rightarrow dx=\frac{1}{t^2}dt \Rightarrow F=\int \frac{dt}{t^2.\frac{1}{t}\sqrt{\frac{1}{t^4}-\frac{1}{t^2}+4}}=\int \frac{dt}{t.\sqrt{\frac{1}{t^4}-\frac{1}{t^2}+4}}=\int \frac{dt}{ \sqrt{t^2.\left [\frac{1}{t^4}-\frac{1}{t^2}+4} \right ]} [/TEX]
[TEX]F =\int \frac{dt}{\sqrt{4t^2+\frac{1}{t^2}+1}}[/TEX]
[TEX]Dat \ t^2=u \Rightarrow \{t=\sqrt{u} \\ \frac{du}{2t}=dt[/TEX]
[TEX] F=\int \frac{du}{2.\sqrt{u}.\sqrt{4u+\frac{1}{u}+1}}= \frac{1}{2} \int \frac{du}{\sqrt{4u^2+u+1}}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn