

InterviewSolution
Saved Bookmarks
1. |
Evaluate:`int1/((x^2+2x+2)^2)dx` |
Answer» `I=(1)/((x^(2)+2x+2)^(2))dx` `=(1)/([(x+1)^(2)+1^(2)]^(2))dx` Put `x+1=tan theta` or `dx=sec^(2)theta d theta` ` :. I= int(1)/((tan^(2)theta+1)^(2))sec^(2)theta d theta` `=int cos^(2)theta d theta` `=(1)/(2)int (1+cos2theta)d theta` `=(1)/(2)(theta+(sin 2theta)/(2))+C ` `=(1)/(2)(theta +sin theta cos theta)+C` `=(1)/(2){tan^(-1)(x+1)+(x+1)/(x^(2)+2x+2)}+C` |
|