g.\(I_7=\int\dfrac{\sqrt{1+x^3}}{x}dx=\int\dfrac{\sqrt{1+x^3}.x^2}{x^3}dx\)Đặt \(\sqrt{1+x^3}=t\Rightarrow x^3=t^2-1\)\(\Rightarrow3x^2dx=2tdt\)\(\Rightarrow x^2dx=\dfrac{2}{3}t.dt\)\(\Rightarrow I_7=\dfrac{2}{3}\int\dfrac{t.tdt}{t^2-1}=\dfrac{2}{3}\int\dfrac{t^2}{t^2-1}dt=\dfrac{2}{3}\int\left(1+\dfrac{1}{t^2-1}\right)dt\)\(=\dfrac{2}{3}\int\left(1+\dfrac{1}{2}.\dfrac{1}{t-1}-\dfrac{1}{2}.\dfrac{1}{t+1}\right)dt\)\(=\dfrac{2}{3}t+\dfrac{1}{3}ln\left|\dfrac{t-1}{t+1}\right|+C\)\(=\dfrac{2}{3}\sqrt{x^3+1}+\dfrac{1}{3}ln\left|\dfrac{\sqrt{x^3+1}-1}{\sqrt{x^3+1}+1}\right|+C\)Câu trả lời: g.\(I_7=\int\dfrac{\sqrt{1+x^3}}{x}dx=\int\dfrac{\sqrt{1+x^3}.x^2}{x^3}dx\)Đặt \(\sqrt{1+x^3}=t\Rightarrow x^3=t^2-1\)\(\Rightarrow3x^2dx=2tdt\)\(\Rightarrow x^2dx=\dfrac{2}{3}t.dt\)\(\Rightarrow I_7=\dfrac{2}{3}\int\dfrac{t.tdt}{t^2-1}=\dfrac{2}{3}\int\dfrac{t^2}{t^2-1}dt=\dfrac{2}{3}\int\left(1+\dfrac{1}{t^2-1}\right)dt\)\(=\dfrac{2}{3}\int\left(1+\dfrac{1}{2}.\dfrac{1}{t-1}-\dfrac{1}{2}.\dfrac{1}{t+1}\right)dt\)\(=\dfrac{2}{3}t+\dfrac{1}{3}ln\left|\dfrac{t-1}{t+1}\right|+C\)\(=\dfrac{2}{3}\sqrt{x^3+1}+\dfrac{1}{3}ln\left|\dfrac{\sqrt{x^3+1}-1}{\sqrt{x^3+1}+1}\right|+C\)