Mathematics

# Find the antiderivative of $f\left( x \right) = \,ln\,\left( {ln\,x} \right) + {\left( {ln\,x} \right)^{ - 2}}$ whose graph passes through $(e, e)$.

##### SOLUTION
$\begin{array}{l} We\, get \\ f\left( x \right) =\ln { \left( { \ln { x } } \right) } +\frac { 1 }{ { { { \left( { \ln { x } } \right) }^{ 2 } } } } \\ \int { f(x)dx } =\int { \left( { \ln { \left( { \ln { x } } \right) } +\frac { 1 }{ { { { \left( { \ln { x } } \right) }^{ 2 } } } } } \right) } dx \\ x={ e^{ t } }\, \, ,dx={ e^{ t } }dt \\ =\int { \left( { \ln { t } +\frac { 1 }{ { { t^{ 2 } } } } } \right) } { e^{ t } }dt \\ =\int { { e^{ t } }\left( { { { lnt } }+\frac { 1 }{ t } } \right) dt } +\int { { e^{ t } }\left( { \frac { 1 }{ { { t^{ 2 } } } } -\frac { 1 }{ t } } \right) dt } \\ =\int { { e^{ t } }\left( { \ln { t } +\frac { { d\left( { \ln { t } } \right) } }{ { dt } } } \right) dt+\int { { e^{ t } }\left( { -\frac { 1 }{ t } +\frac { d }{ { dt } } \left( { \frac { { -1 } }{ t } } \right) dt } \right) } } \\ ={ e^{ t } }\ln { t } -\frac { { { e^{ t } } } }{ t } +C \\ \int { fdx } =x\ln { \left( { \ln { x } } \right) } -\frac { x }{ { \ln { x } } } +C \\ y=x\ln { \left( { \ln { x } } \right) } \frac { { -e } }{ 1 } +c\, \, put\, \, \left( { e,e } \right) \\ e=e\ln { \left( 1 \right) } -\frac { e }{ 1 } +C\, \, \, \, \\ c=2e \\ y=x\ln { \left( { { { lnx } } } \right) } \frac { { -x } }{ { \ln { x } } } +2e \end{array}$

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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 Hard
Evaluate: $\displaystyle\int \sqrt { \dfrac { 1 - x } { 1 + x } } d x$
• A. $\cos^{-1}x+\sqrt{1-x^2}+c$
• B. $\sin^{-1}x-\sqrt{1-x^2}+c$
• C. None of these
• D. $\sin^{-1}x+\sqrt{1-x^2}+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve : $\displaystyle \int \dfrac{5x + 3}{\sqrt{x^2 + 4x + 10}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Evaluate $\displaystyle \int \sqrt {\displaystyle \frac {1\, -\, \sqrt{x}}{1\, +\, \sqrt {x}}} dx$.
• A. $\sqrt{x}\, \sqrt{1\, -\, x}\, -\, 2 \sqrt{1\, -\, x}\, + tan^{-1} \sqrt {x}\, +\, c$
• B. $\sqrt{x}\, \sqrt{1\, -\, x}\, -\, 2 \sqrt{1\, -\, x}\, + sin^{-1} \sqrt {x}\, +\, c$
• C. $\sqrt{x}\, \sqrt{1\, -\, x}\, -\, 2 \sqrt{1\, -\, x}\, + sec^{-1} \sqrt {x}\, +\, c$
• D. $\sqrt{x}\, \sqrt{1\, -\, x}\, -\, 2 \sqrt{1\, -\, x}\, + cos^{-1} \sqrt {x}\, +\, c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate : $\displaystyle \int_{0}^{\pi /2}\sin^3 x\ dx$

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