Монгол Бодлогын Сан

Заавар: $t=\sqrt[3]{x-8}$, $x=t^3+8$ орлуулга ашигла.

Бодолт: $$\begin{align*} \int\dfrac{\sqrt[3]{x-8}}{x} dx&=\left[\begin{array}{c} t=\sqrt[3]{x-8}\\ x=t^3+8\\ dx=3t^2dt\end{array}\right]\\ &=\int\left(\dfrac{t}{t^3+8}\right)3t^2dt=\int\dfrac{3t^3dt}{t^3+8}=\int\left(3-\dfrac{24}{t^3+8}\right)dt\\ &=3t-\int\dfrac{2dt}{t+2}+\int\dfrac{2t-8}{t^2-2t+4}dt\\ &=3t-2\ln|t+2|+\int\dfrac{2(t-1)-6}{(t-1)^2+3}dt\\ &=3t-2\ln|t+2|+\int\dfrac{d[(t-1)^2+3]}{(t-1)^2+3}-\int\dfrac{6dt}{(t-1)^2+(\sqrt{3})^2}\\ &=3t-2\ln|t+2|+\ln|(t-1)^2+3|-\dfrac{6}{\sqrt3}\arctg\dfrac{t-1}{\sqrt3}+C\\ &=3t-2\ln|t+2|+\ln(t^2-2t+4)-\dfrac{6}{\sqrt3}\arctg\dfrac{t-1}{\sqrt3}+C\\ &=3\sqrt[3]{x-8}-2\ln|\sqrt[3]{x-8}+2|+\ln\big((\sqrt[3]{x-8})^3-2\sqrt[3]{x-8}+4\big)-\dfrac{6}{\sqrt3}\arctg\dfrac{\sqrt[3]{x-8}+1}{\sqrt3}+C \end{align*}$$