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If $\int_{0}^{1} \frac{e^{t}}{1 + t} dt = a$ , then $\int_{0}^{1} \frac{e^{t}}{(1 + t)^{2}} dt$ is equal to

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Correct option is B

since, I=∫01 et1+tdt=11+tet01+∫01 et(1+t)2dt=aTherefore, ∫01 et(1+t)2=a−e2+1

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If ∫01 et1+tdt=a , then ∫01 et(1+t)2dt is equal to

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If $\int_{0}^{1} \frac{e^{t}}{1 + t} dt = a$ , then $\int_{0}^{1} \frac{e^{t}}{(1 + t)^{2}} dt$ is equal to