tìm nguyên hàm

S

shibakaovy

$\int\dfrac{1}{x(x^4+1)}.dx$

Đặt $t=x^2$ thì $dt=2xdx$ \Leftrightarrow $\dfrac{dt}{2t}=\dfrac{dx}x$

Vậy

$\int\dfrac{1}{x(x^4+1)}.dx = \int\dfrac{dt}{2t(t^2+1)}=\int(\dfrac1{2t}-\dfrac t{2(t^2+1)})dt=\int(\dfrac1{2t}-\dfrac {2t}{4(t^2+1)})dt=\dfrac12\ln|t|-\dfrac14\ln|t^2+1|=\dfrac12\ln x^2-\dfrac14\ln(x^4+1)$
 
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