tính tích phân

H

hoahongtham_6789

ta có:
I= [TEX]\int_{0}^{1\frac{(x^2+\frac{1}{x^2})+1}{x^4+\frac{1}{x^2}}}[/TEX]dx
đặt t=[TEX]x^2+\frac{1}{x^2}\Rightarrow [/TEX]
 
H

hoahongtham_6789

Nhầm rùi, để coi lại đã,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
 
H

hoanghondo94

[TEX]I=\int_{0}^{1}\frac{x^4+x^2+1}{x^6+1}[/TEX]

em cảm ơn :)
Bấm để xem đầy đủ nội dung ...
Tách nó ra 2 cái đi :p:p:p:p

[TEX]I=\int_{0}^{1}\frac{x^4+x^2+1}{x^6+1}dx = \int_0^1 \frac{x^4 +1}{x^6 + 1}dx +\int_{0}^{1}\frac { x^2}{x^6+1}dx=I_1+I_2[/TEX]



[TEX]I_1 = \int_0^1 \frac{x^4 +1}{x^6 + 1}dx = \int_0^1 \frac{dx}{x^2 + 1}+\frac{1}{3}\int_0^1 \frac{d(x^3)}{(x^3)^2 + 1} \ ---->Okie[/TEX]

[TEX]I_2=\int_{0}^{1}\frac{x^2}{x^6+1}dx=\frac{1}{3} \int \frac{d(x^3)}{x^6+1}=\frac{arctanx^3}{3}[/TEX]
 
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