bai` kho'

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longnhi905

rainshine12 said:
\int_{}^{}(x^5-x)dx/(x^8+1)
cac anh chi giup em cai nha.su?dung nhay tang lau
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đặt t=x2⇒∫x5−xx8+1dx=12∫t2−1t4+1dt=12∫1−1t2t2+1t2dt=12∫d(t+1t)(t+1t)2−2t={x}^{2}\Rightarrow \int \frac{{x}^{5}-x}{{x}^{8}+1}dx =\frac{1}{2}\int \frac{{t}^{2}-1}{{t}^{4}+1}dt = \frac{1}{2}\int \frac{1-\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}}dt = \frac{1}{2}\int \frac{d\left(t+\frac{1}{t} \right)}{{\left(t+\frac{1}{t} \right)}^{2}-2}
đến đây bạn giả ok nha.
 
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