Minkboo

$Có I=\int_{0}^{1}\frac{dx}{(1+x^{n}).\sqrt[n]{1+x^{n}}}=\int_{0}^{1}\frac{1+x^{n}}{(1+x^{n}).\sqrt[n]{1+x^{n}}}dx-\int_{0}^{1}x.\frac{x^{n-1}}{\sqrt[n]{(1+x^{n})^{n+1}}}dx$
$=\int_{0}^{1}\frac{dx}{\sqrt[n]{1+x^{n}}}-\int_{0}^{1}x.\frac{x^{n-1}}{\sqrt[n]{(1+x^{n})^{n+1}}}dx$
Tính $J=\int_{0}^{1}x.\frac{x^{n-1}}{\sqrt[n]{(1+x^{n})^{n+1}}}dx$
Đặt $\left\{\begin{matrix} u=x & \\ v'=\frac{x^{n-1}}{\sqrt[n]{(1+x^{n})^{n+1}}} & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} u'=1 & \\ v=-\frac{1}{\sqrt[n]{1+x^{n}}} & \end{matrix}\right.$
$\Rightarrow J=-\frac{x}{\sqrt[n]{1+x^{n}}}\mid _{0}^{1}+\int_{0}^{1}\frac{dx}{\sqrt[n]{1+x^{n}}}$
Vậy $I=\int_{0}^{1}\frac{dx}{\sqrt[n]{1+x^{n}}}dx-J=\frac{x}{\sqrt[n]{1+x^{n}}}\mid _{0}^{1}=\frac{1}{\sqrt[n]{2}}$

Có $\int \frac{x.e^{x}+1}{(e^{x}+x)^{2}}dx=\int \frac{e^{x}(e^{x}+x)-e^{2x}+1}{(e^{x}+x)^{2}}dx=\int \frac{e^{x}}{e^{x}+x}dx-\int \frac{e^{2x}-1}{(e^{x}+x)^{2}}dx$
Tính $I=\int \frac{e^{x}}{e^{x}+x}dx$
$\left\{\begin{matrix}
Đặt u=\frac{1}{e^{x}+x} & \\
v'=e^{x}&
\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
u'=-\frac{e^{x}+1}{(e^{x}+x)^{2}} & \\
v=e^{x} &
\end{matrix}\right.$
$\Rightarrow I=\frac{e^{x}}{e^{x}+x}+\int \frac{e^{2x}+e^{x}}{(e^{x}+x)^{2}}dx$
Vậy $\int \frac{x.e^{x}+1}{(e^{x}+x)^{2}}dx=\frac{e^{x}}{e^{x}+x}+\int \frac{e^{2x}+e^{x}}{(e^{x}+x)^{2}}dx-\int \frac{e^{2x}-1}{(e^{x}+x)^{2}}dx$
$=\frac{e^{x}}{e^{x}+x}+\int \frac{e^{x}+1}{(e^{x}+x)^{2}}dx=\frac{e^{x}}{e^{x}+x}-\frac{1}{e^{x}+x}=\frac{e^{x}-1}{e^{x}+x}$

Bài 32:
Có $\int_{4}^{3}\frac{x^{10}}{x^{3}+1}dx=\int_{4}^{3}(x^{7}-x^{4}+x-\frac{1}{x^{2}-x+1}+\frac{1}{x^{3}+1})dx$
$=\int_{4}^{3}(x^{7}-x^{4}+x-\frac{1}{(x-\frac{1}{2}^{2})+(\frac{\sqrt{3}}{2})^{2}}+\frac{1}{x^{3}+1})dx$
Tính $I=\int_{4}^{3}\frac{1}{x^{3}+1}dx =\int_{4}^{3}\frac{d(x+1)}{(x+1)[x^{2}-x+1]}=\int_{4}^{3}\frac{d(x+1)}{(x+1)[(x+1)^{2}-3(x+1)+3]}$
Đặt $t=x+1$ $\Rightarrow dt=dx$
Đổi cận $x:4\rightarrow 3 \Rightarrow t:5\rightarrow 4$
$\Rightarrow I=\frac{1}{3}\int_{5}^{4}\frac{(t^{2}-3t+3)-(t^{2}-3t)}{t(t^{ 2}-3t+3)}dt$
$=\frac{1}{3}\int_{5}^{4}\frac{dt}{t}-\frac{1}{3}\int_{5}^{4}\frac{t-3}{t^{2}-3t+3}dt$
$=\frac{1}{3}\int_{5}^{4}\frac{dt}{t}-\frac{1}{3}(\frac{1}{2}\int_{5}^{4}\frac{2t-3}{t^{2}-3t+3}dt-\frac{3}{2}\int_{5}^{4}\frac{dt}{t^{2}-3t+3})=...$
Vậy $\int_{4}^{3}=\frac{x^{10}}{x^{3}+1}dx=\int_{4}^{3}(x^{7}-x^{4}+x-\frac{1}{(x-\frac{1}{2}^{2})+(\frac{\sqrt{3}}{2})^{2}})dx+I=...$