Mathematics

# Solve : $\int \dfrac{x^4dx}{(1+x^2)^3}$.

##### SOLUTION
Given the integral,
$\int { \dfrac { { x }^{ 4 }dx }{ { ({ x }^{ 2 }+1) }^{ 3 } } }$
using partial fraction we get,
$=\int { \left( \dfrac { 1 }{ { x }^{ 2 }+1 } -\dfrac { 2 }{ { ({ x }^{ 2 }+1) }^{ 2 } } +\dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } } \right) } \\ =\int { \dfrac { 1 }{ { x }^{ 2 }+1 } } dx-2\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } } } dx+\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } } } dx$
Here,
$\int { \dfrac { 1 }{ { x }^{ 2 }+1 } } dx\\ =\tan^{-1} { (x) }$
For,
$\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } } } dx$
applying reduction formula we get,
$=\dfrac { x }{ 2({ x }^{ 2 }+1) } +\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { x }^{ 2 }+1 } } dx\\ =\dfrac { \tan^{-1} { (x) } }{ 2 } +\dfrac { x }{ 2({ x }^{ 2 }+1) }$
Again for,
$\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } } } dx$
applying reduction formula we get,
$=\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } } +\dfrac { 3 }{ 4 } \int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } } } dx\\ =\dfrac { 3\tan^{-1} { (x) } }{ 8 } +\dfrac { 3x }{ 8({ x }^{ 2 }+1) } +\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } } \\ \therefore \int { \dfrac { 1 }{ { x }^{ 2 }+1 } } dx-2\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } } } dx+\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } } } dx\\ =\dfrac { 3\tan^{-1} { (x) } }{ 8 } -\dfrac { 5x }{ 8({ x }^{ 2 }+1) } +\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } }$
Hence,
$\int { \dfrac { { x }^{ 4 }dx }{ { ({ x }^{ 2 }+1) }^{ 3 } } } \\ =\dfrac { 3\tan^{-1} { (x) } }{ 8 } -\dfrac { 5x }{ 8({ x }^{ 2 }+1) } +\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } } +C\\ =\dfrac { 3\tan^{-1} { (x) } }{ 8 } +\dfrac { -5{ x }^{ 3 }-3x }{ 8{ ({ x }^{ 2 }+1) }^{ 2 } } +C.$

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

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle f\left ( x \right )=\int_{-1}^{1}\frac{\sin x}{1+t^{2}}dt$ then $\displaystyle {f}'\left ( \frac{\pi }{3} \right )$ is
• A. nonexistent
• B. $\displaystyle \pi \sqrt{3/4}$
• C. none of these
• D. $\displaystyle \pi /4$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int \dfrac{px^{p+2q-1}-qx^{q-1}}{x^{2p+2q}+2x^{p+q}+1}dx$ is equal to

• A. $-\dfrac{x^{p}}{x^{p+q}+1}+C$
• B. $\dfrac{x^{q}}{x^{p+q}+1}+C$
• C. $\dfrac{x^{p}}{x^{p+q}+1}+C$
• D. $-\dfrac{x^{q}}{x^{p+q}+1}+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Integrate $\displaystyle \int_{1}^{2}(x^2-1)dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Matrix Hard
The value of $\displaystyle \int_{a}^{b}f\left ( x \right )dx$
 $a= 0,\:b= \pi /3,\:f\left ( x \right )= \displaystyle \frac{x}{1+\sin x}$ $\displaystyle \frac{16}{3}\pi -2\sqrt{3}$ $a= 1,\:b= 16,\:f\left ( x \right )= \tan^{-1}\sqrt{\sqrt{x}-1}$ $\displaystyle \frac{\pi ^{2}}{16}+\displaystyle \frac{\pi }{4} \log 2$ $a= 0,\:b= \pi /2,\:f\left ( x \right )= \displaystyle \frac{x\sin 2x}{\left ( \sin x+\cos x \right )^{2}}$ $-\displaystyle \frac{\pi }{3} \left ( 2-\sqrt{3} \right )+\log \left ( \dfrac{1}{2}+\displaystyle \frac{\sqrt{3}}{2} \right )$ $a= 0,\:b= \pi /4,\:f\left ( x \right )= \displaystyle \frac{x^{2}\sec ^{2}x\left ( \cos 2x-\sin 2x \right )}{\left ( \sin x+\cos x \right )^{2}}$ $\displaystyle \frac{\pi ^{2}}{8}-\displaystyle \frac{\pi }{4}$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { 2x }{ 1-{ x }^{ 2 } } \right) dx } } =\dfrac{\pi}{a}-\ln a$. Find $a$.
• A. $1$
• B. $-1$
• C. None of these
• D. $2$