Mathematics

# $\displaystyle \int_{0}^{\infty }\frac{x}{\left ( 1+x \right )\left ( 1+x^{2} \right )}dx$

$\displaystyle \frac{\pi }{4}$

is sme as $\displaystyle \int_{0}^{\infty }\frac{dx}{\left ( 1+x \right )\left ( 1+x^{2} \right )}$

##### SOLUTION
Let $I=\int _{ 0 }^{ \infty }{ \cfrac { xdx }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } }$
Using partial fraction
$=\int _{ 0 }^{ \infty }{ \left( \cfrac { x+1 }{ 2\left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2\left( 1+x \right) } \right) dx } \\ ={ \left( \lim _{ b\rightarrow \infty }{ \cfrac { 1 }{ 2 } \log { \left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+x \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }x } \right) }_{ 0 }^{ b }\\ =\lim _{ b\rightarrow \infty }{ \left( \cfrac { 1 }{ 2 } \log { \left( 1+b \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+b \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }b \right) } \\ =\cfrac { \pi }{ 4 }$

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

#### Realted Questions

Q1 Subjective Medium
Obtain: $\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \cfrac { 1 }{ 1-x+{ x }^{ 2 } } \right) } } dx$, where $0< x< 1$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
Evaluate:$\displaystyle\int \frac{dx}{\sqrt{1+4x^{2}}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int\limits_0^\frac \pi 2 x\sqrt {1-x^2} dx =$
• A. $\dfrac{\pi }{4}$
• B. $\dfrac{\pi }{6}$
• C. $\dfrac{\pi}{8}$
• D. $\dfrac{\pi }{3}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Integrate with respect to $x$.:
$e^{x}\sin x$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$4 \displaystyle \int \dfrac{\sqrt{a^6 + x^8}}{x} dx$ is equal to ______________________.
• A. $a^6 ln\vert \dfrac {\sqrt{a^6 + x^8} - a^3} {\sqrt {a^6 + x^8} + a^3}\vert + c$
• B. $\sqrt{a^6 + x^8} + \dfrac {a^3}{2} ln \vert \dfrac {\sqrt{a^6 + x^8} - a^3} {\sqrt {a^6 + x^8} + a^3}\vert + c$
• C. $a^6 ln \vert \dfrac {\sqrt{a^6 + x^8} + a^3} {\sqrt {a^6 + x^8} - a^3}\vert + c$
• D. $\sqrt{a^6 + x^8} + \dfrac{a^3}{2} ln \vert \dfrac {\sqrt{a^6 + x^8} + a^3} {\sqrt {a^6 + x^8} - a^3}\vert + c$