Mathematics

# Evaluate: $\displaystyle\int {\left( {1 - \dfrac{{\log \dfrac{1}{v}}}{{{v^2}}}} \right)} \,dv$

##### SOLUTION

Consider given the given integration,

$I=\int{1-\dfrac{\log \dfrac{1}{v}}{{{v}^{2}}}dv}$

$I=\int{1dv-\int{\dfrac{\log \dfrac{1}{v}}{{{v}^{2}}}dv}}=v-\int{\dfrac{\log \dfrac{1}{v}}{{{v}^{2}}}dv}$        …..(1)

Let,

$y=\int{\dfrac{\log \dfrac{1}{v}}{{{v}^{2}}}dv}$

Put,

$t=\dfrac{1}{v}$

$dt=\dfrac{-1}{{{v}^{2}}}dv$

$dv=-{{v}^{2}}dt$

$y=\int{\dfrac{-{{v}^{2}}\log t}{{{v}^{2}}}dv}=-\int{\left( \log t \right).1dt}$

$=-\left[ \log t.t-\int{\dfrac{1}{t}.tdt} \right]$

$=-t.\log t+t+C$

$y=-\dfrac{1}{{{v}^{2}}}\log \dfrac{1}{{{v}^{2}}}+\dfrac{1}{{{v}^{2}}}+C$

Put, the value of  in equation (1),we get

$I=v-\left( -\dfrac{1}{{{v}^{2}}}\log \dfrac{1}{{{v}^{2}}}+\dfrac{1}{{{v}^{2}}}+C \right)$

$I=v+\dfrac{1}{{{v}^{2}}}\log \dfrac{1}{{{v}^{2}}}-\dfrac{1}{{{v}^{2}}}-C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle {I}_{ {n}}=\int^{\pi/2 }_{\pi/4}( {T} {a} {n}\theta)^{- {n}}. {d}\theta$ for $( {n}>1)$
then $I_{n}+I_{n+2} = ?$
• A. $\displaystyle \frac{-1}{\mathrm{n}+1}$
• B. $\displaystyle \frac{1}{\mathrm{n}-1}$
• C. $\displaystyle \frac{-1}{\mathrm{n}-1}$
• D. $\displaystyle \frac{1}{\mathrm{n}+1}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
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Q3 Subjective Medium
Integrate:
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Q4 Subjective Medium
Find the value of $\displaystyle\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}|\sin x|\ dx$.

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