Đặt $u=\sqrt[3]{e^x-1}\\
\Rightarrow u^3=e^x-1\\
\Rightarrow 3u^2du=e^xdx$
Đổi cận
$x=0\Rightarrow u=0\\
x=ln2\Rightarrow u=1$
Nên $I=\int_{0}^{1}\dfrac{3u^3}{u^3+1}du\\
=\int_{0}^{1}3du-\int_{0}^{1}\dfrac{3}{(u+1)(u^2-u+1)}du\\
=3u-I_1\\
=3-I_1$
Ta có $\dfrac{3}{(u+1)(u^2-u+1)}=\dfrac{A}{u+1}+\dfrac{Bu+C}{u^2-u+1}$
Nên $3=A(u^2-u+1)+(u+1)(Bu+C)=(A+B)u^2+(B+C-A)u+A+C$
Ta có hệ
$ \begin{cases} A+B=0 \\ B+C-A=0 \\ A+C=3 \end{cases}\\
\Leftrightarrow \begin{cases}A=1\\ B=-1\\ C=2 \end{cases}$
Nên $I_1=\int_{0}^{1}\dfrac{du}{u+1}+
\int_{0}^{1}\dfrac{-u+2}{u^2-u+1}du\\
=ln(u+1)-\int_{0}^{1}\dfrac{2u-1}{2(u^2-u+1)}du+\dfrac{3}{2}\int_{0}^{1}\dfrac{1}{(u-\dfrac{1}{2})^2+\dfrac{3}{4}}du\\
=ln2-\dfrac{1}{2}ln(u^2-u+1)+\dfrac{3}{2}I_2\\
= ln2+\dfrac{3}{2}I_2$
Đặt $\dfrac{\sqrt{3}}{2}tant=u-\dfrac{1}{2}$
$\Leftrightarrow \dfrac{\sqrt{3}}{2}(1+tan^2t)dt=du$
Nên $I_2=\int_ {-\dfrac{\pi}{6}}^{\dfrac{\pi}{6}}\dfrac{2}{\sqrt{3}}dt$
$=\dfrac{2t}{\sqrt{3}}\\
=\dfrac{2\pi}{3\sqrt{3}}$
Vậy $I=3-ln2-\dfrac{\pi}{\sqrt{3}}$
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