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If I1=∫ee2d...

Question

If $I_{1} = \int_{e}^{e^{2}} \frac{d x}{ln x}$ and $I_{2} = \int_{1}^{2} \frac{e^{x}}{x} d x$ , then

A

I_{1} = I_{2}

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B

2 I_{1} = I_{2}

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C

I_{1} = 2 I_{2}

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D

2 I_{1} = 3 I_{2}

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I_{1} = \int_{e}^{e^{2}} \frac{d x}{l o g x}

I_{2} = \int_{1}^{2} \frac{e^{x}}{x} d x,

I_{1} = \int_{1}^{2} x (\sqrt{x} + \sqrt{3 - x}) d x

I_{2} = \int_{1}^{2} (\sqrt{x} + \sqrt{3 - x}) d x

\frac{I_{1}}{I_{2}} =

I_{1} = \int_{0}^{1} \frac{{tan}^{- 1} x d x}{x}

I_{2} = \int_{0}^{\frac{π}{2}} \frac{x d x}{sin x}

\frac{I_{1}}{I_{2}}

I_{1} = \int_{0}^{π / 2} \frac{x}{sin x} d x

I_{2} = \int_{0}^{π / 2} \frac{{tan}^{- 1} x}{x} d x,

\frac{I_{1}}{I_{2}} =

f (x) = \frac{e^{x}}{1 + e^{x}}

I_{1} = \int_{f (- a)}^{f (a) x g {x (1 - x)}} d x

I_{2} = \int_{f (- a)}^{f (a) g {x (1 - x)}} d x

\frac{I_{2}}{I_{1}}

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