-1e1-x ln x - -ln xx dx equals

Name: JEE- Grade 12- Mathematics- Definite Integration- Session 02- D06
Uploaded: 2022-07-04T10:36:42+05:30
Description: I =∫_1^e1 + ln x√(x ln x) dx put xln x = t^2 ⇒ (ln x + 1) dx = 2t dt I =∫_0^√(e)2t dtt = 2 √(e) ...

I =∫_1^e1 + ln x√(x ln x) dx put xln x = t^2 ⇒ (ln x + 1) dx = 2t dt I =∫_0^√(e)2t dtt = 2 √(e) ...

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Byju's Answer

Standard XII

Mathematics

Sandwich Theorem

∫1e1√x ln x +...

Question

$e \int 1 (\frac{1}{\sqrt{x ln x}} + \sqrt{\frac{ln x}{x}}) d x$ equals

\sqrt{e}

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2 e

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2 \sqrt{e}

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\sqrt{2 e}

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Solution

The correct option is C $2 \sqrt{e}$
$I = e \int 1 \frac{1 + ln x}{\sqrt{x ln x}} d x$
put $x ln x = t^{2} \Rightarrow (ln x + 1) d x = 2 t d t$
$I = \sqrt{e} \int 0 \frac{2 t d t}{t} = 2 \sqrt{e}$

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Sandwich Theorem

Standard XII Mathematics

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e∫1(1√x ln x+√lnxx)dx equals

The correct option is C 2√eI=e∫11+lnx√xlnxdx put xlnx=t2⇒(lnx+1)dx=2tdt I=√e∫02tdtt=2√e

$e \int 1 (\frac{1}{\sqrt{x ln x}} + \sqrt{\frac{ln x}{x}}) d x$ equals

The correct option is C $2 \sqrt{e}$
$I = e \int 1 \frac{1 + ln x}{\sqrt{x ln x}} d x$
put $x ln x = t^{2} \Rightarrow (ln x + 1) d x = 2 t d t$
$I = \sqrt{e} \int 0 \frac{2 t d t}{t} = 2 \sqrt{e}$