$\int\frac{x^5}{x^2+1}dx=\int(x^3-x+\frac{x}{x^2+1})dx$
$=\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2}ln(x^2+1)+C$
$\Rightarrow I=[\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2}ln(1+x^2)]|^1_0=\frac{2ln2-1}{4}$
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