First slide

Introduction to integration

Question

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

Moderate

A

$a - 1 + \frac{e}{2}$
B

$a + 1 - \frac{e}{2}$
C

$a - 1 - \frac{e}{2}$
D

$a + 1 + \frac{e}{2}$

Solution

since, $I = \int_{0}^{1} \frac{e^{t}}{1 + t} dt = {[\frac{1}{1 + t} e^{t}]}_{0}^{1} + \int_{0}^{1} \frac{e^{t}}{(1 + t)^{2}} dt = a$
Therefore, $\int_{0}^{1} \frac{e^{t}}{(1 + t)^{2}} = a - \frac{e}{2} + 1$

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