Mathematics

# Solve :$\displaystyle \int \left (\log(\log x)+\dfrac{1}{(\log x)^2}\right)dx$

##### SOLUTION
$I=\displaystyle\int { \left( \log(\log x)+\dfrac { 1 }{ { (\log x) }^{ 2 } } \right) dx } \\ \quad =\displaystyle\int { \log(\log x)dx } +\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ Let\quad { I }_{ 1 }=\displaystyle\int { \log(\log x)dx } \\ \quad \quad \quad\quad =\displaystyle\int { \log(\log x).1dx } \\ \text{continue with integration by part}:\\ \displaystyle\int { f(x)g(x)dx } =f(x)\displaystyle\int { g(x)dx } -\int { \left[ f'(x)\int { g(x)dx } \right] } dx\\ now,\\ { { I }_{ 1 } }=\log(\log x)\displaystyle\int { 1 } dx\quad -\quad \int { \left[ \dfrac { d }{ dx } (\log(\log x)).\int { 1dx } \right] } dx\\ \quad \quad =x\log(\log x)\quad -\quad \displaystyle\int { \left[ \dfrac { 1 }{ \log x } .\dfrac { 1 }{ x } .x \right] } dx\\ \quad \quad =x\log(\log x)\quad -\quad\displaystyle \int { \dfrac { 1 }{ \log x } } dx\\ Let\quad { I }_{ 2 }=\displaystyle\int { \dfrac { 1 }{ \log x } } dx\\ \quad \quad \quad \quad \quad =\displaystyle\int { \dfrac { 1 }{ \log x } .1dx } \\ \quad \quad \quad \quad \quad =\dfrac { 1 }{ \log x } .\displaystyle\int { 1dx } \quad -\quad \int { \left[ \dfrac { d }{ dx } \left( \dfrac { 1 }{ \log x } \right) .\displaystyle\int { 1dx } \right] } dx\\ \quad \quad \quad \quad \quad =\quad \dfrac { x }{ \log x } \quad -\quad \displaystyle\int { \dfrac { -1 }{ { (\log x) }^{ 2 } } .\dfrac { 1 }{ x } .xdx } \\ \quad \quad \quad \quad \quad =\dfrac { x }{ \log x } +\displaystyle\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ putting\quad the\quad value\quad of\quad { I }_{ 2 }\quad in\quad { I }_{ 1 }\\ { I }_{ 1 }=x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad -\quad \displaystyle\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ now\quad putting\quad the\quad value\quad of\quad { I }_{ 1 }\quad in\quad I\\ I\quad =\quad x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad -\quad\displaystyle \int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \quad +\quad \int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ \quad \quad =\quad x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad +\quad C,\quad where\quad C\quad =\quad integration\quad constant\quad$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
Evaluate the following integral
$\int { \cfrac { \sec ^{ 2 }{ x } }{ \tan { x } +2 } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\lim\limits_{n\to \infty}\dfrac{1^2+2^2+.....+n^2}{n^3}$ is equal to
• A. $\infty$
• B. $0$
• C. $\dfrac{1}{2}$
• D. $\dfrac{1}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium

If $\displaystyle \int_{0}^{60}\frac{dx}{2x+1}=\log a$, then $a=$
• A. $3$
• B. $81$
• C. $40$
• D. $11$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
$\int_0^{\tfrac{2}{3}} {\dfrac{{dx}}{{4 + 9{x^2}}} = }$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$