∫etan−1x(1+x+x2).d(cot−1x) is equal to
−etan−1x+C
etan−1x+C
−x.etan−1x+C
x.etan−1x+C
I=∫etan−1x(1+x+x2).(−11+x2)dx =−∫etan−1x(1+x1+x2)dx =−∫etan−1xdx−∫x.etan−1x1+x2dx =−∫etan−1xdx−x.etan−1x+∫etan−1xdx+C =−x.etan−1x+C