Mathematics

# $\displaystyle \int \dfrac{dx}{(1+x^2)^2}$

##### SOLUTION
$Let,\quad I=\int { \dfrac {\ dx }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } } } \\ we\quad will\quad use\quad the\quad substitution\quad x=\tan \theta \\ \Rightarrow \ dx={ \sec }^ { 2 }\theta d\theta \\ =\int { \dfrac { { \sec }^ { 2 }\theta d\theta }{ { \sec }^ { 4 }\theta } } \quad \quad \left[ \because 1+{ \tan }^ { 2 }\theta ={ \sec }^ { 2 }\theta \right] \\ =\int { \dfrac { d\theta }{ { \sec }^ { 2 }\theta } } \\ =\int { { \cos }^ { 2 }\theta d\theta } \\ =\dfrac { 1 }{ 2 } \int { \cos 2\theta d\theta +\int { \dfrac { 1 }{ 2 } d\theta } } \quad \quad \left[ \because \cos 2\theta =2{ \cos }^ { 2 }\theta -1\Rightarrow { \cos }^ { 2 }\theta =\dfrac { 1 }{ 2 } \cos 2\theta +\dfrac { 1 }{ 2 } \right] \\ =\dfrac { 1 }{ 4 } \sin 2\theta +\dfrac { 1 }{ 2 } \theta +C\\$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 TRUE/FALSE Medium
State whether true or false:

$\displaystyle \int _{ 0 }^{ \pi /2 }\left( \cos { x } \right) dx = 1$
• A. False
• B. True

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int\frac{x\cos x\log x-\sin x}{x(\log x)^{2}}dx=$
• A. $\displaystyle \frac { \cos { x } }{ \log { x } } +C$
• B. $\displaystyle \frac { \log { x } }{ \sin { x } } +C$
• C. $\displaystyle \frac { \log { x } }{ \cos { x } } +C$
• D. $\displaystyle \frac { \sin { x } }{ \log { x } } +C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
$\displaystyle\int^{\pi}_0\dfrac{x\sin 2x\sin\left(\dfrac{\pi}{2}\cos x\right)}{2x-\pi}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int cosec^{n}x\cdot cotxdx,(n\neq 0)=$
• A. $\displaystyle \dfrac{cosec^{n}x}{n}+c$
• B. $\dfrac{cosec^{n+1}x}{n+1}+c$
• C. $-\displaystyle \dfrac{cosec^{n+1}x}{n+1}+c$
• D. $-\displaystyle \dfrac{cosec^{n}x}{n}+c$

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 $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$