Mathematics

# Resolve into partial fractions $\displaystyle \frac{x^2+2}{(x+1)^3(x-2)}$

$\displaystyle - \frac{6}{(x+2)}+\frac{6}{(x+1)}-\frac{5}{(x+1)^2}+\frac{3}{(x+1)^3}$

##### SOLUTION
Let $\displaystyle \frac { x^{ 2 }+2 }{ (x+1)^{ 3 }(x-2) } =\frac { A }{ \left( x+1 \right) } +\frac { B }{ { \left( x+1 \right) }^{ 2 } } +\frac { C }{ { \left( x+1 \right) }^{ 3 } } +\frac { D }{ \left( x-2 \right) }$
$\Rightarrow \left( { x }^{ 2 }+2 \right) =A{ \left( x+1 \right) }^{ 2 }\left( x-2 \right) +B\left( x+1 \right) \left( x-2 \right) +C\left( x-2 \right) +D{ \left( x+1 \right) }^{ 3 }$
$\Rightarrow { x }^{ 2 }+2=A\left( { x }^{ 3 }-x-2 \right) +B\left( { x }^{ 2 }-x-2 \right) +C\left( x-2 \right) +D\left( { x }^{ 3 }+3{ x }^{ 2 }+3x+1 \right)$
On comparing we get
$A+D=0,B+D=1,-A-B+C+3=0,-2A-2B-2C+D=2\\ \Rightarrow A=6,B=-5,C=3,D=-6$
Hence
$\displaystyle \frac { x^{ 2 }+2 }{ (x+1)^{ 3 }(x-2) } =\frac { 6 }{ \left( x+1 \right) } -\frac { 5 }{ { \left( x+1 \right) }^{ 2 } } +\frac { 3 }{ { \left( x+1 \right) }^{ 3 } } -\frac { 6 }{ \left( x-2 \right) }$
Hence, option 'B' is correct.

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

#### Realted Questions

Q1 Subjective Medium
$\text { Evaluate } \int_{0}^{\pi} e^{2 x} \cdot \sin \left(\dfrac{\pi}{4}+x\right) \mathrm{d} \mathrm{x}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle\int \dfrac{dx}{x^3(1+x^6)^{2/3}}=x$. $f(x)\cdot (1+x^6)^{1/3}+C$, then $f(x)$ is equal to?
• A. $\dfrac{-1}{2x^2}$
• B. $\dfrac{-1}{6x^2}$
• C. $\dfrac{1}{6x^2}$
• D. $\dfrac{-1}{2x^3}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $y=\displaystyle \int \displaystyle \frac{dx}{(1+x^2)^{{3}/{2}}}$ and  $y=0$ when $x=0$, then value of $y$ when $x=1$ is
• A. $\sqrt{\displaystyle \frac{2}{3}}$
• B. $\sqrt{2}$
• C. $\displaystyle \frac{1}{2}$
• D. $\displaystyle \frac{1}{\sqrt{2}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Prove that $\displaystyle\int^{\pi/2}_0(2 log \cos x-log \sin 2x)dx=-\dfrac{\pi}{2}(log 2)$.

$\displaystyle \int_{0}^{\pi /2} \sin 2x\ dx$