$\int\limits_{-1}^{1}\frac{x^4+sinx}{x^2+1}dx$
có đáp an rồi kia –  Wade 04-01-15 10:59 PM
30p r -_- –  Minn 04-01-15 10:55 PM
5p nưa nha –  Wade 04-01-15 10:42 PM
m chỉ t đi :)) –  Minn 04-01-15 10:35 PM
:)) hahaa –  Wade 04-01-15 10:34 PM
t suýt làm đc –  Minn 04-01-15 10:32 PM
$I=\int_1^{-1} \dfrac{x^4}{x^2+1} dx + \int_1^{-1} \dfrac{\sin x}{x^2+1}dx=I_1 +I_2$

Nói chung chả ai viết cận dở hơi như đứa ra đề. Tóm lại $I_2=0$ do cận dối, hàm dưới dấu tích phân là hàm lẻ

Còn $I_1 = -2\int_{0}^1 \dfrac{x^4}{x^2 +1}dx$ đơn giản
nghiếm câm a giải bài kia nha –  Wade 04-01-15 11:10 PM
với a thì cái đó là nhảm nhí :D –  Dép Lê Con Nhà Quê 04-01-15 11:07 PM
cài này ok –  thuylinhnguyen2611 04-01-15 11:01 PM
tuy a ơi a phải cmt tại sao nó bằng ko chứ –  Wade 04-01-15 10:58 PM

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