Mathematics

# $\int\limits_0^\infty {\dfrac{{dx}}{{({x^2} + {a^2})({x^2} + {b^2})}}} =$

##### SOLUTION
$\dfrac{1}{\left ( x^{2}+a^{2} \right )+\left ( x^{2}+b^{2} \right )} = \dfrac{1}{b^{2}-a^{2}}\left ( \dfrac{1}{\left ( x^{2}+a^{2} \right )}- \dfrac{1}{\left ( x^{2}+b^{2} \right )} \right )$

$So,\int_{0}^{\infty }\left ( \dfrac{1}{\left ( x^{2}+a^{2} \right )+\left ( x^{2}+b^{2} \right )} \right )dx= \dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty } \dfrac{1}{\left ( x^{2}+a^{2} \right )}dx \right )-\dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty } \dfrac{1}{\left ( x^{2}+b^{2} \right )}dx \right )$

$As \int \dfrac{1}{\left ( x^{2}+a^{2} \right )dx}=\\\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$
$So,\dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty } \dfrac{1}{\left ( x^{2}+a^{2} \right )}dx \right )-\dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty } \dfrac{1}{\left ( x^{2}+b^{2} \right )}dx \right ) = \\\\\dfrac{1}{b^{2}-a^{2}}\times \dfrac{1}{a}(\tan^{-1}\dfrac{x}{a})_{0}^{\infty } -\dfrac{1}{b^{2}-a^{2}} \times \dfrac{1}{b}(\tan^{-1}\dfrac{x}{b})_{0}^{\infty }$
After putting limits , we will get
$= \dfrac{\pi}{2(b^{2}-a^{2})}\left ( \dfrac{1}{a}-\dfrac{1}{b} \right )$

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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
$\displaystyle\int^{\dfrac{\pi}{4}}_{\dfrac{-\pi}{4}}\sqrt{\dfrac{1-\cos 2008x}{2}}dx$ equals?
• A. $\dfrac{1}{251}$
• B. $\dfrac{1}{498}$
• C. $\dfrac{1}{502}$
• D. $\dfrac{1}{249}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate : $\displaystyle\int^2_0\sqrt{6x+4}dx$
• A. $\dfrac{64}{9}$
• B. $7$
• C. $\dfrac{60}{9}$
• D. $\dfrac{56}{9}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\displaystyle\int^8_1\dfrac{dx}{x^{2/3}}$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle\int { \cfrac { 1 }{ 5+4\sin { x } } } dx=A\tan ^{ -1 }{ \left( B\tan { \cfrac { x }{ 2 } } +\cfrac { 4 }{ 3 } \right) } +C\quad$, then
• A. $A=\cfrac { 1 }{ 3 } ,B=\cfrac { 2 }{ 3 }$
• B. $A=-\cfrac { 2 }{ 3 } ,B=\cfrac { 5 }{ 3 }$
• C. $A=\cfrac { 1 }{ 3 } ,B=-\cfrac { 5 }{ 3 }$
• D. $A=\cfrac { 2 }{ 3 } ,B=\cfrac { 5 }{ 3 }$

$\int { \cfrac { f'(x) }{ f(x) } dx } =\log { [f(x)] } +c$