Answer
$$\int^{e}_1\frac{1}{x}(1+7\ln x)^{-1/3}dx=\frac{9}{14}$$
Work Step by Step
$$A=\int^{e}_1\frac{1}{x}(1+7\ln x)^{-1/3}dx$$
We set $u=1+7\ln x$, which means $$du=\frac{7}{x}dx$$ $$\frac{1}{x}dx=\frac{1}{7}du$$
For $x=e$, we have $$u=1+7\ln e=1+7=8$$
For $x=1$, we have $$u=1+7\ln1=1+7\times0=1$$
Therefore, $$A=\frac{1}{7}\int^8_1u^{-1/3}du$$ $$A=\frac{1}{7}\times\frac{3u^{2/3}}{2}\Big]^8_1=\frac{3}{14}u^{2/3}\Big]^8_1$$ $$A=\frac{3}{14}(8^{2/3}-1)$$ $$A=\frac{3}{14}(4-1)=\frac{9}{14}$$
