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=$-\dfrac{1}{10}\int (\dfrac{1}{1+x^{10}})΄x^{5}dx$=$-\dfrac{x^{5}}{10(1+x^{10})}+\dfrac{1}{10}\int \dfrac{5x^{4}}{1+x^{10}}dx$=(u=x^{5})$-\dfrac{x^{5}}{10(1+x^{10})}+\dfrac{1}{10}\int \dfrac{1}{1+u^{2}}du$=$-\dfrac{x^{5}}{10(1+x^{10})}+\dfrac{arctanx^{5}}{10}+c$
1 σχόλιο:
=$-\dfrac{1}{10}\int (\dfrac{1}{1+x^{10}})΄x^{5}dx$=
ΑπάντησηΔιαγραφή$-\dfrac{x^{5}}{10(1+x^{10})}+\dfrac{1}{10}\int \dfrac{5x^{4}}{1+x^{10}}dx$=(u=x^{5})
$-\dfrac{x^{5}}{10(1+x^{10})}+\dfrac{1}{10}\int \dfrac{1}{1+u^{2}}du$=
$-\dfrac{x^{5}}{10(1+x^{10})}+\dfrac{arctanx^{5}}{10}+c$