giải giúp bài tích phân khó

H

hoanghondo94

vannhan1345 said:
\int_{}^{}dx/(x^4+1)

giải chi tiết vào cho dễ hiểu

:khi (55)::khi (55):
Bấm để xem đầy đủ nội dung ...


[TEX]{\color{Blue} I=\int \frac{dx}{x^4+1}=\frac{1}{2}\int \frac{(x^2+1)-(x^2-1)}{x^4+1}dx=\frac{1}{2}\left ( \int \frac{x^2+1}{x^4+1}dx-\frac{x^2-1}{x^4+1}dx \right ) \\\\ =\frac{1}{2}\left ( \int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx- \int \frac{1+\frac{1}{x^2}}{x^2-\frac{1}{x^2}}dx\right )=\frac{1}{2}\left ( \int \frac{d(x-\frac{1}{x})}{(x-\frac{1}{x})^2+(\sqrt{2})^2} -\int \frac{d(x+\frac{1}{x})}{(x+\frac{1}{x})^2-(\sqrt{2})^2}\right ) \\\\ =\frac{1}{2}\left ( \frac{1}{\sqrt{2}}arctan\frac{x^2-1}{x\sqrt{2}}-\frac{1}{2\sqrt{2}}ln\left | \frac{x^2-x\sqrt{2}+1}{x^2+x\sqrt{2}+1} \right | \right )+C[/TEX];););););)
 
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