∫ 0 1 2 5 2 − x d x = 2 5 ∫ 0 1 1 2 − x d x = 2 5 ( ln | 2 − x | ( 1 − 1 ) ) | 0 1 = − 2 5 ( ln | 2 − 1 | − ln | 2 − 0 | ) . {\displaystyle \int _{0}^{1}{\frac {\frac {2}{5}}{2-x}}\,dx={\frac {2}{5}}\int _{0}^{1}{\frac {1}{2-x}}\,dx={\frac {2}{5}}\left.\left(\ln |2-x|\left({\frac {1}{-1}}\right)\right)\right|_{0}^{1}={\frac {-2}{5}}(\ln |2-1|-\ln |2-0|).} Don't forget the extra factor 1/(-1) which comes by taking the derivative of the inner function (2-x). Same with the second integral which needs an extra factor of 1/2 since (1+2x)' = 2. Does this make it more clear?
Ooooh! Okay, thanks. I didn't even know that was a thing that needed to be done!
Don't forget the extra factor 1/(-1) which comes by taking the derivative of the inner function (2-x). Same with the second integral which needs an extra factor of 1/2 since (1+2x)' = 2. Does this make it more clear?
