Tích phân hàm phân thức

D

ducpro1990

V

vodichhocmai

[TEX]I=\int_{0}^{2}\frac{x^2-1}{1+x^4}dx=\int_0^2\frac{1}{\frac{4x^2}{\(1+x^2\)^2}-2}\ \frac{2(1-x^2)}{\(1+x^2\)^2}dx[/TEX]

[TEX]t=\frac{2x}{1+x^2}\rightarrow{I=\int_0^{\frac{4}{5}}\frac{1}{t^2-2}dt=\frac{1}{2\sqrt2}ln\|\frac{t-\sqrt2}{t+\sqrt2}\|\|_0^{\frac{4}{5}}[/TEX]
 
T

tungkoy

kinh nhi tach duoc nhu the cung la kho tuong tuong nhi cha nhe k co cach nao de hieu hon ah may anh chi


kimxakiem2507:

[TEX]*\int_0^2\frac{x^2-1}{x^4+1}dx=\int_0^2\frac{x^2-1}{(x^2+1)^2-2x^2}dx=\int_0^2[\frac{\frac{\sqrt2}{2}x-\frac{1}{2}}{x^2-\sqrt2x+1}-\frac{\frac{\sqrt2}{2}x+\frac{1}{2}}{x^2+\sqrt2x+1}]dx=\frac{1}{2\sqrt2}ln\frac{x^2-\sqrt2x+1}{x^2+\sqrt2x+1}\|_0^2[/TEX]
 
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