Mathematics

# $\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+1 } } dx=$

$\dfrac { 1 }{ \sqrt { 2 } } {\tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$

##### SOLUTION
Given,

$\int \dfrac{x^2+1}{x^4+1}dx$

$=\int \dfrac{\dfrac{1}{x^2}+1}{\dfrac{1}{x^2}+x^2}dx$..............dividing by ${x^2}$

$=\int \dfrac{\dfrac{1}{x^2}+1}{\dfrac{1}{x^2}+x^2-2+2}dx$

$=\int \dfrac{\dfrac{1}{x^2}+1}{\left ( x-\dfrac{1}{x} \right )^2+2}dx$

$x-\dfrac{1}{x}=t\rightarrow dt=\dfrac{1}{x^2}+1 dx$

$=\int \dfrac{1}{t^2+2}dt$

$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{t}{\sqrt 2}+c$

$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x-\dfrac{1}{x}}{\sqrt 2}+c$

$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x^2-1}{\sqrt 2x}+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Subjective Medium
Solve:
$\displaystyle\int {\dfrac{{4x + 6}}{{2{x^2} + 5x + 3}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the following integrals:
$\int { \cfrac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int \frac{dx}{x^{2}+2x+2}dx$  is equal to
• A. $(x+1)\tan^{-1}x+c$
• B. $x \tan ^{-1}(x+1)+c$
• C. $\tan^{-1}x+c$
• D. $\tan ^{-1}(x+1)+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\int ^{\pi /4}_{-\pi /4} (\cos 3x-\sin 3x)^{2}dx=$
• A. $\displaystyle \frac{\pi}{4}$
• B. $2\pi$
• C. $-2\pi$
• D. $\displaystyle \frac{\pi}{2}$

$\displaystyle \int\limits_{0}^{\pi/2} \cos^2x \ dx$