Tìm tích phân :
[tex]I_{1}=\int_{0}^{1}\frac{x^{2}}{(x+1)^{3}}dx[/tex]
Đặt $u=x^2\Rightarrow du=2xdx$
$dv=(x+1)^{-3}dx\Rightarrow v=-\dfrac{1}{2}(x+1)^{-2}$
$I_1=\displaystyle\int\limits_0^1\dfrac{x^2}{(x+1)^3}dx=-\dfrac{x^2}{2(x+1)^2}\left|\begin{matrix}1\\0\end{matrix}\right.+\displaystyle\int\limits_0^1\dfrac{xdx}{(x+1)^2}$
Đặt $u=x\Rightarrow du=dx$
$dv=(x+1)^{-2}dx\Rightarrow v=-(x+1)^{-1}$
$\displaystyle\int\limits_0^1\dfrac{xdx}{x+1}=-\dfrac{x}{x+1}\left|\begin{matrix}1\\0\end{matrix}\right.+\displaystyle\int\limits_0^1\dfrac{dx}{x+1}$
$\displaystyle\int\limits_0^1\dfrac{dx}{x+1}=\ln|x+1|\left|\begin{matrix}1\\0\end{matrix}\right.$
Suy ra $I_1=-\dfrac{x^2}{2(x+1)^2}\left|\begin{matrix}1\\0\end{matrix}\right.-\dfrac{x}{x+1}\left|\begin{matrix}1\\0\end{matrix}\right.+\ln|x+1|\left|\begin{matrix}1\\0\end{matrix}\right.=-\dfrac{1}{8}-\dfrac{1}{2}+\ln2=-\dfrac{5}{8}+\ln2$
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