Let I -Q1 e x -1- x dx-01 x - x -12 e x dx equalsA- e -2- IB- I - e -2- e -2-1- ID- I -1- e -2

The correct option is A e/2 I ∫_0^1(x/x+1 )^2 e^x dx =∫ e^x+1/(x+1)^2e^x dx 2I =e/2 I ...

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Algebra of Derivatives

Let I =∫Q1 e ...

Question

Let $I = \int_{0}^{1} \frac{e^{x}}{1 + x} d x$
$\int_{0}^{1} {(\frac{x}{x + 1})}^{2} e^{x} d x$ equals

\frac{e}{2} - I

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I - \frac{e}{2}

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

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

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Solution

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Similar questions

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\int_{3}^{4} \frac{e^{- x}}{x - 5}

I

I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x, I_{2} \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x

I_{3} = \int_{0}^{1} e^{- x^{2}} d x, I_{4} = \int_{0}^{1} e^{- x^{2} / 2} d x,

I_{1} = \int \frac{d x}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}

I_{2} = \int \frac{d x}{e^{3 x} + 8 e^{x} + 4 e^{- x}}

I = I_{1} - 2 I_{2}

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