Đáp án: $ I=x-2-9\sqrt[3]{x-2}+9\sqrt{3}\arctan \left(\dfrac{\sqrt[3]{x-2}}{\sqrt{3}}\right)$
Giải thích các bước giải:
Đặt $\sqrt[3]{x-2}=t$
$\to d(\sqrt[3]{x-2})=dt$
$\to (\sqrt[3]{x-2})'dx=dt$
$\to \dfrac{1}{3\sqrt[3]{(x-2)^2}}dx=dt$
$\to \dfrac{1}{3t^2}dx=dt$
$\to dx=3t^2dt$
Khi đó:
$I=\int\dfrac{t^2}{3+t^2}\cdot 3t^2dt$
$\to I=\int\dfrac{3t^4}{3+t^2}dt$
$\to I=\int\dfrac{t^4}{1+(\dfrac{t}{\sqrt{3}})^2}dt$
Đặt $\dfrac{t}{\sqrt{3}}=y$
$\to t=\sqrt{3}y$
$\to dt=\sqrt{3}dy$
$\to I=\int\dfrac{(\sqrt{3}y)^4}{1+y^2}\sqrt{3}dy$
$\to I=\int 9\sqrt{3}\cdot \dfrac{y^4}{1+y^2}dy$
$\to I= 9\sqrt{3}\cdot\int \dfrac{y^4}{1+y^2}dy$
$\to I= 9\sqrt{3}\cdot\int \dfrac{y^4-1+1}{1+y^2}dy$
$\to I= 9\sqrt{3}\cdot\int \dfrac{(y^2-1)(y^2+1)+1}{1+y^2}dy$
$\to I= 9\sqrt{3}\cdot\int y^2-1 \dfrac{1}{1+y^2}dy$
$\to I= 9\sqrt{3}\cdot ( \dfrac{y^3}{3}-y+\arctan y)$
$\to I= 9\sqrt{3}\cdot ( \dfrac{(\dfrac{t}{\sqrt{3}})^3}{3}-\dfrac{t}{\sqrt{3}}+\arctan(\dfrac{t}{\sqrt{3}}))$
$\to I=t^3-9t+9\sqrt{3}\arctan \left(\dfrac{t}{\sqrt{3}}\right)$
$\to I=(\sqrt[3]{x-2})^3-9\sqrt[3]{x-2}+9\sqrt{3}\arctan \left(\dfrac{\sqrt[3]{x-2}}{\sqrt{3}}\right)$
$\to I=x-2-9\sqrt[3]{x-2}+9\sqrt{3}\arctan \left(\dfrac{\sqrt[3]{x-2}}{\sqrt{3}}\right)$
