This discussion is now closed.
Scroll to see replies
\displaystyle \int \dfrac {x}{x+1}\\[br][br][br]= \displaystyle \int \dfrac {x+1-1}{x+1} \\[br][br][br]= \displaystyle \int \dfrac {x+1}{x+1} - \dfrac {1}{x+1}\\[br][br][br]= x - \ln (x+1)[br]
\displaystyle\int \dfrac {x}{x+1}\\\\[br][br][br]= \displaystyle\int \dfrac {x+1-1}{x+1}\\\\[br][br][br]= \displaystyle\int \dfrac {x+1}{x+1} - \dfrac {1}{x+1}\\\\[br][br][br]= x - \ln (x+1)[br]
\pi\displaystyle\int x^{2} dy \\[br]