Mathematics

# Resolve into partial fractions $\displaystyle \frac{4x^3-7x^2+5x-4}{(x^2+1)(x^2-3x+2)}$

$\displaystyle \frac{x}{x^2+1}+\frac{2}{x-2}+\frac{1}{x-1}$

##### SOLUTION
Let $\displaystyle \frac { 4x^{ 3 }-7x^{ 2 }+5x-4 }{ \left( x^{ 2 }+1 \right) \left( x^{ 2 }-3x+2 \right) } =\frac { Ax+B }{ { x }^{ 2 }+1 } +\frac { C }{ x-2 } +\frac { D }{ x-1 }$
$\Rightarrow 4x^{ 3 }-7x^{ 2 }+5x-4=\left( Ax+B \right) \left( x-2 \right) \left( x-1 \right) +C\left( { x }^{ 2 }+1 \right) \left( x-1 \right) +D\left( { x }^{ 2 }+1 \right) \left( x-2 \right) \\ \Rightarrow 4x^{ 3 }-7x^{ 2 }+5x-4=A\left( { x }^{ 3 }-3{ x }^{ 2 }+2x \right) +B\left( { x }^{ 2 }-3x+2 \right) +C\left( { x }^{ 3 }-{ x }^{ 2 }-x-1 \right) +D\left( { x }^{ 3 }-2{ x }^{ 2 }+x-2 \right)$
On comparing coefficients we get
$A+C+D=4,-3A+B-C-2D=-7,2A-3B-C+D=5,2B-C-2D=-4\\ \Rightarrow A=1,B=0,C=2,D=1$
Hence
$\displaystyle \frac { 4x^{ 3 }-7x^{ 2 }+5x-4 }{ \left( x^{ 2 }+1 \right) \left( x^{ 2 }-3x+2 \right) } =\frac { x }{ { x }^{ 2 }+1 } +\frac { 2 }{ x-2 } +\frac { 1 }{ x-1 }$
Hence, option 'C' is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Hard
Evaluate: $\int (1 - x)\sqrt {x}dx.$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve : $\displaystyle \int \dfrac{x^2 + x + 5}{3x + 2} dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate the given integral.
$\displaystyle\quad \int { \cfrac { x }{ ({ x }^{ 2 }-{ a }^{ 2 })({ x }^{ 2 }-{ b }^{ 2 }) } } dx\quad$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\int^3_1 x^2dx$ is:
• A. $\dfrac{28}{3}$
• B. $\dfrac{25}{3}$
• C. None of these
• D. $\dfrac{26}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts
$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
where  $\displaystyle v_{1}\left ( x \right )=\int v\left ( x \right )dx,v_{2}\left ( x \right )=\int v_{1}\left ( x \right )dx ..., v_{n}\left ( x \right )= \int v_{n-1}\left ( x \right )dx$
Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.