Answer
$$\int^{9}_{0}\frac{2\log_{10}(x+1)}{x+1}dx=\ln10$$
Work Step by Step
$$A=\int^{9}_{0}\frac{2\log_{10}(x+1)}{x+1}dx=2\int^{9}_{0}\frac{\frac{\ln(x+1)}{\ln10}}{x+1}dx$$ $$A=\frac{2}{\ln10}\int^{9}_{0}\frac{\ln(x+1)}{x+1}dx$$
We set $u=\ln(x+1)$, which means $$du=\frac{(x+1)'}{x+1}dx=\frac{1}{x+1}dx$$
- For $x=9$, we have $u=\ln10$
- For $x=0$, we have $u=\ln1=0$
Therefore, $$A=\frac{2}{\ln10}\int^{\ln10}_0udu=\frac{2}{\ln10}\times\frac{u^2}{2}\Big]^{\ln10}_0$$ $$A=\frac{u^2}{\ln10}\Big]^{\ln10}_0$$ $$A=\frac{(\ln10)^2-0^2}{\ln10}$$ $$A=\frac{(\ln10)^2}{\ln10}=\ln10$$
