If \(\displaystyle\int_0^1 \dfrac{e^t}{1+t}dt=a\), then \(\displayst

1.	If \(\displaystyle\int_0^1 \dfrac{e^t}{1+t}dt=a\), then \(\displaystyle\int_0^1 \dfrac{e^t}{(1+t)^2}dt=\)1. \(a-1+\dfrac{e}{2}\)2. \(a+1+\dfrac{e}{2}\)3. \(a-1-\dfrac{e}{2}\)4. \(a+1-\dfrac{e}{2}\)
Answer» Correct Answer - Option 4 : \(a+1-\dfrac{e}{2}\) Explanation: Given Integral is, \(\displaystyle∫_0^1 \dfrac{e^t}{1+t}dt=a\) Considering \(I_1 ~=~\frac{1}{1+t}\) and \(I_2 ~=~e^t \) Hence by Integration by parts we know that, ∫ [I₁][ I₂] dt = I₁∫ I₂ dt - ∫ [I'₁∫ I₂dt]dt \(\frac{1}{1+t}\int_0^1 e^t dt-\int_0 ^1 [\frac{d}{dt}(\frac{1}{1+t}).\int_0 ^1 e^t dt]dt~=~a\) \(\frac{1}{1+t}[e^t ]_0^1-\int_0^1 (\frac{-1}{(1+t)^2}).e^t dt~=~a\) \([\frac{e^t}{1+t}]_0^1 +\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a\) \(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^t}{1+t}]_0^1\) \(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^1}{1+1}-\frac{e^0}{1+0}]\) \(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e}{2}-1]\) \(\displaystyle\int_0^1 \dfrac{e^t}{(1+t)^2}dt~=~a+1-\frac{e}{2}\)

If \(\displaystyle\int_0^1 \dfrac{e^t}{1+t}dt=a\), then \(\displaystyle\int_0^1 \dfrac{e^t}{(1+t)^2}dt=\)1. \(a-1+\dfrac{e}{2}\)2. \(a+1+\dfrac{e}{2}\)3. \(a-1-\dfrac{e}{2}\)4. \(a+1-\dfrac{e}{2}\)

Answer» Correct Answer - Option 4 : \(a+1-\dfrac{e}{2}\)

Explanation:

Given Integral is,

\(\displaystyle∫_0^1 \dfrac{e^t}{1+t}dt=a\)

Considering \(I_1 ~=~\frac{1}{1+t}\) and \(I_2 ~=~e^t \)

Hence by Integration by parts we know that,

∫ [I₁][ I₂] dt = I₁∫ I₂ dt - ∫ [I'₁∫ I₂dt]dt

\(\frac{1}{1+t}\int_0^1 e^t dt-\int_0 ^1 [\frac{d}{dt}(\frac{1}{1+t}).\int_0 ^1 e^t dt]dt~=~a\)

\(\frac{1}{1+t}[e^t ]_0^1-\int_0^1 (\frac{-1}{(1+t)^2}).e^t dt~=~a\)

\([\frac{e^t}{1+t}]_0^1 +\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^t}{1+t}]_0^1\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^1}{1+1}-\frac{e^0}{1+0}]\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e}{2}-1]\)

\(\displaystyle\int_0^1 \dfrac{e^t}{(1+t)^2}dt~=~a+1-\frac{e}{2}\)