∫1+2x+3x2+4x3+−−−−dx where 0<|x|<1
11−x+c
23(1−x)2+c
4(1−x)3+c
(1−x)+c
let
S=1+2x+3x2+4x3+−−−xS= x+2x2+3x3+−−−(1−x)S=1+x+x2+x3+−−−(1−x)S=11−xS=1(1−x)2S=(1−x)−2∴∫1+2x+3x2+−−−−dx=∫(1−x)−2dx=(1−x)−1(−1)(−1)+c=(1−x)−1+c