\int (e^x-\sqrt x+ \frac{1}{x}+1) dx
S Stud778 New member Joined Jun 10, 2010 Messages 3 Jun 10, 2010 #1 ∫(ex−x+1x+1)dx\displaystyle \int (e^x-\sqrt x+ \frac{1}{x}+1) dx∫(ex−x+x1+1)dx
D Deleted member 4993 Guest Jun 10, 2010 #3 Stud778 said: ∫(ex−x+1x+1)dx\displaystyle \int (e^x-\sqrt x+ \frac{1}{x}+1) dx∫(ex−x+x1+1)dx Click to expand... ∫(ex−x+1x+1)dx = ∫exdx − ∫xdx + ∫1xdx + ∫1dx\displaystyle \int (e^x-\sqrt x+ \frac{1}{x}+1) dx \ \ = \ \ \int e^x dx \ \ - \ \ \int \sqrt {x} dx \ \ + \ \ \int \frac{1}{x} dx \ \ + \ \ \int 1dx∫(ex−x+x1+1)dx = ∫exdx − ∫xdx + ∫x1dx + ∫1dx
Stud778 said: ∫(ex−x+1x+1)dx\displaystyle \int (e^x-\sqrt x+ \frac{1}{x}+1) dx∫(ex−x+x1+1)dx Click to expand... ∫(ex−x+1x+1)dx = ∫exdx − ∫xdx + ∫1xdx + ∫1dx\displaystyle \int (e^x-\sqrt x+ \frac{1}{x}+1) dx \ \ = \ \ \int e^x dx \ \ - \ \ \int \sqrt {x} dx \ \ + \ \ \int \frac{1}{x} dx \ \ + \ \ \int 1dx∫(ex−x+x1+1)dx = ∫exdx − ∫xdx + ∫x1dx + ∫1dx
