tích phân ^_^

D

dien0709

$I=\int \dfrac{x^3}{1+\sqrt[4]{1+x^4}}$

$u=\sqrt[4]{1+x^4}\to u^4=1+x^4\to u^3du=x^3dx$

$I=\int \dfrac{u^3du}{1+u}=\int \dfrac{u^3+1-1}{u+1}du$

$I=\int (u^2-u+1-\dfrac{1}{1+u})du$ dễ rồi
 
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