If \int_{0}^{1} \frac{e^x}{1+x} \dx=\alpha, then ...

CBSE Class 12 Math 2025 All Sets Solved Paper

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Question : 3 of 20

Marks: +1, -0

If

∫↙{0}↖{1} \;{e^x}/{1+x} \dx=α,

then

∫↙{0}↖{1} \;{e^x}/{(1+x)^2} \dx

is equal to

$α-1+\;{e}/{2}$
$α+1-\;{e}/{2}$
$α-1-\;{e}/{2}$
$α+1+\;{e}/{2}$

Solution:

∫ ↙{0}↖{1} \;{e^{x}}/{1+x} \dx=α

Here we will apply integration by parts formula
Which is

∫ u v \dx=u ∫ v \dx-∫ (u^{'} ∫ v \dx) \dx

Here

u=\;{1}/{1+x}

and

v=e^{x}

⇒ u^{'}=\;{-1}/{(1+x)^{2}}

∫↙{0}↖{1} \;{e^{x}}/{1+x} \dx= [{1}{(1+x)}/{e}^{x}] _{0}^{1}-∫ ↙{0}↖{1} (\;{-1}/{(1+x)^{2}}) e^{x} \dx

α=\;{e}/{2}-1+∫ ↙{0}↖{1} \;{e^{x}}/{1+x^{2}} \dx

⇒ ∫↙{0}↖{1} \;{e^{x}}/{(1+x)^{2}} \dx=α+1-\;{e}/{2}

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