If -01 et dt-1 - t - A then the value of -01 et dt-1 - t2

The correct option is D A e/2 + 1 ∫_0^1e^t(1+t)^2dt Let u=e^t⇒ du=e^tdt dv=1(1+t)^2dt⇒ v= 11+t = [e^t1+t]_0^1+∫_0^1e^t1+tdt ...

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Standard XII

Mathematics

Property 1

If ∫01 et d...

Question

If $1 \int 0 \frac{e^{t} d t}{1 + t} =$ A then the value of $1 \int 0 \frac{e^{t} d t}{{(1 + t)}^{2}}$

A + \frac{e}{2} - 1

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A - \frac{e}{2} + 1

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A - \frac{e}{2} - 1

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A + \frac{e}{2} + 1

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Solution

The correct option is D $A - \frac{e}{2} + 1$
$\int_{0}^{1} \frac{e^{t}}{{(1 + t)}^{2}} d t$
Let $u = e^{t} \Rightarrow d u = e^{t} d t$
$d v = \frac{1}{{(1 + t)}^{2}} d t \Rightarrow v = - \frac{1}{1 + t}$
$= - {[\frac{e^{t}}{1 + t}]}_{0}^{1} + \int_{0}^{1} \frac{e^{t}}{1 + t} d t$
$= - [\frac{e}{2} - \frac{1}{1}] + \int_{0}^{1} \frac{e^{t}}{1 + t} d t$
$= - \frac{e}{2} + 1 + A$
$= A - \frac{e}{2} + 1$

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If 1∫0etdt1+t= A then the value of 1∫0etdt(1+t)2

The correct option is D A−e2+1∫10et(1+t)2dtLet u=et⇒du=etdtdv=1(1+t)2dt⇒v=−11+t=−[et1+t]10+∫10et1+tdt=−[e2−11]+∫10et1+tdt=−e2+1+A=A−e2+1

If $1 \int 0 \frac{e^{t} d t}{1 + t} =$ A then the value of $1 \int 0 \frac{e^{t} d t}{{(1 + t)}^{2}}$