Mathematics

# Integrate:$\int{\dfrac{dx}{{{x}^{5}}\left( 1+{{x}^{-4}} \right)}}$

##### SOLUTION

Consider the given integral.

$I=\int{\dfrac{dx}{{{x}^{5}}\left( 1+{{x}^{-4}} \right)}}$

Let $t={{x}^{-4}}$

$\dfrac{dt}{dx}=-4{{x}^{-5}}$

$-\dfrac{dt}{4}=\dfrac{dx}{{{x}^{5}}}$

Therefore,

$I=-\dfrac{1}{4}\int{\dfrac{dt}{\left( 1+t \right)}}$

$I=-\dfrac{1}{4}\ln \left( 1+t \right)+C$

On putting the value of $t$, we get

$I=-\dfrac{1}{4}\ln \left( 1+{{x}^{-4}} \right)+C$

$I=-\dfrac{1}{4}\ln \left( 1+\dfrac{1}{{{x}^{4}}} \right)+C$

$I=-\dfrac{1}{4}\ln \left( \dfrac{{{x}^{4}}+1}{{{x}^{4}}} \right)+C$

$I=\dfrac{1}{4}\ln \left( \dfrac{{{x}^{4}}}{{{x}^{4}}+1} \right)+C$

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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 Hard
$\int { \cfrac { dx }{ \sqrt { \left( 1-{ e }^{ 2x } \right) } } } =\quad$?
• A. $\log { \left| { e }^{ x }+\sqrt { { e }^{ 2x }-1 } \right| } +C\quad$
• B. $\log { \left| { e }^{ -x }+\sqrt { { e }^{ -2x }-1 } \right| } +C$
• C. none of these
• D. $-\log { \left| { e }^{ -x }+\sqrt { { e }^{ -2x }-1 } \right| } +C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle\int\dfrac{e^x}{x}(1+x\log |x|)\ dx=$
• A. $e^x\log|x|+e^x+c$
• B. $e^x\log|s|-x+c$
• C. none of these
• D. $e^x\log |x|+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate $\displaystyle \int \frac{(x - 1)^{2}}{x^{4} + 2x^{2} + 1} dx$
• A. $\displaystyle \frac{x^{3}}{3} + x + \frac{x}{x^{2} + 1} + c$
• B. $\displaystyle \frac{x^{5} + x^{3} + x + 3}{3(x^{2} + 1)} + c$
• C. $\displaystyle \frac{x^{5} + 4x^{3} + 3x + 3}{3(x^{2} + 1)} + c$
• D. None of these

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integrals
$\int { \cfrac { 1 }{ p+q\tan { x } } } dx\quad$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$