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a
11−x+c
b
23(1−x)2+c
c
4(1−x)3+c
d
(1−x)+c
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Correct option is A
letS=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+cSimilar Questions
The value of is
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