If I-01et-1-t dt- then p-01etlog 1-t dt-

The correct option is C elog2 I Given, nbsp; nbsp;p=∫_0^1e^tlog ( 1+t )dt Also,given, I=∫_0^1e^t/1+t dt I= [e^ t log nbsp;( 1+t ) ] _ 0 ...

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Logarithmic Differentiation

If I=∫01et/...

Question

If $I = \int_{0}^{1} \frac{e^{t}}{1 + t}$ dt, then $p = \int_{0}^{1} e^{t} log (1 + t) d t =$

I

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2 I

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e log 2 - I

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none

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A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{- 1 + i \sqrt{3}}{2 i} & \frac{- 1 - i \sqrt{3}}{2 i} \frac{1 + i \sqrt{3}}{2 i} & \frac{1 - i \sqrt{3}}{2 i} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦, i = \sqrt{- i}

f (x) = x^{2} + 2

f (A)

1 \int 0 \frac{e^{t}}{1 + t} d t = a,

1 \int 0 \frac{e^{t}}{(1 + t)^{2}}

z = (1 + i) (1 - i \sqrt{3}) (- 2 - 2 i) (i) (3),

I (m, n) = \int_{- 0}^{1} t^{m} (1 + t)^{n} d t

I (m, n)

I (m + 1, n - 1)

I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{⊖}

\frac{- d [I_{2}]}{d t} = - \frac{1}{2} \frac{d [S_{2} O_{3}^{2 -}]}{d t}

\frac{- d [I_{2}]}{d t} = - 2 \frac{d [S_{2} O_{3}^{2 -}]}{d t}

\frac{- d [I_{2}]}{d t} = - 2 \frac{d [I^{⊖}]}{d t} \frac{d [S_{2} O_{3}^{2 -}]}{d t}

\frac{d [S_{4} O_{6}^{2 -}]}{d t} = \frac{1}{2} \frac{d [I^{⊖}]}{d t}

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