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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^1 \frac{ e ^{-2 x}}{1+ e ^{-x}} dx =$
  • A
    $\log \left(\frac{1+e}{e}\right)-\frac{1}{e}+1$
  • ✓
    $\log \left(\frac{1+e}{2 e}\right)-\frac{1}{e}+1$
  • C
    $\log \left(\frac{1+e}{2 e}\right)+\frac{1}{e}-1$
  • D
    $\log \left(\frac{1+e}{2}\right)+\frac{1}{e}-1$
✓

Answer

Correct option: B.
$\log \left(\frac{1+e}{2 e}\right)-\frac{1}{e}+1$
(B)
Put $1+ e ^{-x}= t \Rightarrow- e ^{-x} d x= dt$
When $x=0, t =2$ and when $x=1, t =1+\frac{1}{ e }$
$\therefore \quad \int_0^1 \frac{ e ^{-2 x}}{1+ e ^{-x}} d x=\int_2^{1+\frac{1}{ e }} \frac{( t -1)(- dt )}{ t }=\int_2^{1+\frac{1}{ e }}\left(\frac{1}{ t }-1\right) dt$
$\begin{array}{l}=[\log t - t ]_2^{1+\frac{1}{ e }} \\ =\log \left(1+\frac{1}{ e }\right)-\left(1+\frac{1}{ e }\right)-\log 2+2 \\ =\log \left(\frac{ e +1}{2 e }\right)-\frac{1}{ e }+1\end{array}$

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