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Q.

∫1+2x+3x2+4x3+−−−−dx where 0<|x|<1

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a

11−x+c

b

23(1−x)2+c

c

4(1−x)3+c

d

(1−x)+c

answer is A.

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Detailed Solution

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+c

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