Question

Hello! Could someone help me with integral, please?

Answer 1

$\displaystyle \int \ln(x)dx\implies \int 1\cdot \ln(x)dx \\\\[-0.35em] ~\dotfill\\\\ \displaystyle u=\ln(x)\implies \cfrac{du}{dx}=\cfrac{1}{x}\hspace{5em}v=\int 1\cdot dx\implies v=x \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \int \ln(x)dx\implies x\ln(x)-\int\cfrac{1}{x}\cdot x\implies x\ln(x)-x+C \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \left. \cfrac{}{}x\ln(x)-x \right]_{2}^{8}\implies [8\ln(8)-8] -[2\ln(2)-2]\implies 8\ln(8)-2\ln(2)-6$

Answer 2

Refer to the photo taken.