Question

$\large\underline{\bold{Given \:Question - }}$

See Above ​

Answer 1

$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \large\mathfrak{\dag \: Solution :- }\:\:\:\:\:\:\:\:\:\:\:\:\:\:$

Formula :-

• $\sf\red { \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}$

$\sf \: Here, \: we \: are \: asked \: to \:resolve \: into \: partial \: fraction : \: \dfrac{x}{ {x}^{3} + 1}$

$\sf\: \dfrac{x}{ {x}^{3} + 1} = \sf \: \dfrac{x}{(x + 1)( {x}^{2} - x + 1) }$

$\sf \: Let \: \: \red { \dfrac{x}{(x + 1)( {x}^{2} - x + 1) } = \sf \: \dfrac{a}{x + 1} + \dfrac{bx + c}{ {x}^{2} - x + 1 } } - - (1)$

$\sf \: x \: = a( {x}^{2} - x + 1) + (bx + c)(x + 1) - - (2)$

$\sf :\implies \: - 1 = a(1 + 1 + 1) + 0$

$\sf :\implies \: - 1 = 3a$

$\sf:\implies \red {\:a \: = \: - \: \dfrac{1}{3} - - (3)}$

❍On substituting 'x = - 0' in equation (2), we get :-

$\sf :\implies\: 0 = a(0 - 0 + 1) + c(0 + 1)$

$\sf :\implies \:a + c = 0$

$\sf :\implies\:c = - \: a$

$\sf:\implies \red {\:\:c \: = \: \dfrac{1}{3} - - (4)}$

❍On substituting 'x = 1' in equation (2), we get :-

$\sf :\implies \:1 = a(1 - 1 + 1) + (b + c)(2)$

$\sf :\implies \:1 = a + 2b + 2c$

$\sf :\implies \:1 = - \: \dfrac{1}{3} + \dfrac{2}{3} + 2b$

$\sf :\implies \:1 = \dfrac{1}{3} + 2b$

$\sf :\implies \:2b = 1 - \dfrac{1}{3}$

$\sf :\implies \:2b = \dfrac{2}{3}$

$\sf:\implies \red {\:b \: = \: \dfrac{1}{3} \: - - (5)}$

❍Now, substitute the values..

$\sf \: \: \dfrac{x}{(x + 1)( {x}^{2} - x + 1) } = \sf \: \dfrac{ - 1}{3(x + 1)} + \dfrac{x + 1}{3( {x}^{2} - x + 1) }$

$\sf \: \: \purple{\dfrac{x}{ {x}^{3} + 1 } } = \sf \pink {\: \dfrac{ - 1}{3(x + 1)} + \dfrac{x + 1}{3( {x}^{2} - x + 1) }}$

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$\sf \green {Do \: you \: know \:?¿}$

• Partial fraction decomposition is the breaking down of a rational expression into simpler parts. It is the opposite of adding rational expressions. When adding two rational expressions, there has to be a common denominator.

$\begin{gathered}\boxed{\begin{array}{c|c} \sf \pink { Term \: in \: denominator }& \sf\pink { Partial \: fraction \: decomposition }\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \sf \dfrac{A}{{ax + b}} \\ \\ \sf {(ax + b)}^{2} & \sf \displaystyle \sf \dfrac{{{A_1}}}{{ax + b}} + \sf\frac{{{A_2}}}{{{{\sf\left( {ax + b} \right)}^2}}} \\ \\ \sf {ax}^{2} + bx + c & \sf \displaystyle \sf \dfrac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}$

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Answer 2

To Do :-

• To resolve $\sf\dfrac{x}{x^3+1}$ into partial fractions .

SolutioN :-

The process of expressing a proper fraction as a sum of two or more proper fraction is called resolving it into Partial Fractions.

Here the fraction given to us is ,

$\sf\to \pink{\dfrac{x}{x^3+1}}$

Here the degree of denominator is greater than the degree of numerator . Hence the fraction is Proper Fraction and we can resolve it into Partial fractions .

• Assume that ,

$\dashrightarrow \sf \dfrac{x}{(x+1)(x^2 - x + 1 )}= \dfrac{A}{x+1} + \dfrac{Bx + C }{x^2-x+1} \\\\\sf\dashrightarrow \dfrac{x}{(x+1)(x^2 - x + 1 )}= \dfrac{A(x^2-x+1)+ (Bx+C)(x+1)}{(x^2-x+1)(x+1)} \\\\ \qquad\qquad\tiny{\underline{\dag\red{ \sf On \ comparing \ we \ have }}} \\\\\sf\dashrightarrow x = A(x^2-x+1) +( Bx+C)(x+1) \\\\\sf\qquad\qquad\tiny{\underline{\dag\red{ \sf Substituting \ x \ = \ (-1) . }}} \\\\\sf\dashrightarrow -1 = A[ (-1)^2-(-1)+1] + (Bx+C)(x-1) \\\\\sf\dashrightarrow -1 = A( 1+1+1)+ 0 \\\\\sf\dashrightarrow -1 = 3A \\\\\sf\dashrightarrow\boxed{\pink{\sf A =\dfrac{-1}{3}}}$

$\rule{200}2$

$\qquad\qquad \tiny{\underline{\dag\red{ \sf Substituting \ x \ = \ (0) . }}} \\\\\sf\dashrightarrow 0 = A( 0 -0+1)+[ (B)(0)+ C ] \\\\\sf\dashrightarrow0 = A + C \\\\\sf\dashrightarrow \dfrac{-1}{3}+C = 0 \qquad\bigg\lgroup \red{\tt From \ above }\bigg\rgroup \\\\\sf\dashrightarrow \boxed{\pink{\sf C =\dfrac{1}{3}}}$

$\rule{200}2$

Comparing the co - efficients of x² on both sides of the equation we have ,

$\sf\dashrightarrow A + B = 0 \\\\\sf\dashrightarrow \dfrac{-1}{3}+B = 0 \\\\\sf\dashrightarrow \boxed{\pink{\sf B =\dfrac{1}{3}}}$

$\rule{200}2$

$\red{\bigstar}\underline{\textsf{ Put on the respective values :- }}$

$\sf \to \dfrac{A}{x+1} + \dfrac{Bx + C }{x^2-x+1} \\\\\sf\to \dfrac{\frac{-1}{3}}{x+1} + \dfrac{\frac{1}{3}x+\frac{1}{3}}{x^2-x+1} \\\\\sf\to\dfrac{-1}{3(x+1)} + \dfrac{\frac{1+x}{3}}{x^2-x+1} \\\\\sf\to \underset{\blue{\sf Required \ Partial \ Fraction }}{\underbrace{\boxed{\pink{\sf{\dfrac{-1}{3(x+1)}+\dfrac{1+x}{3(x^2-x+1)}}}}} }$

$\rule{200}2$

$\large\red{\bigstar}\underline{\textsf{ More To Know :- }}$

Let us assume a fraction ( proper ) , $\sf \dfrac{f(x)}{g(x)}$

T Y P E - 1 :-

When g(x) has non repeated linear fractions only , then there exists a corresponding partial fraction in the form of

$\to \sf\red{ \dfrac{A}{ax+b}}$

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T Y P E - 2 :-

When g(x) has some repeated linear factors and the remaining are non repeated linear factors .

$\sf Let \ g(x) = (px+q)^n(a_{1}x+a_{1})(a_{2}x+a_2).. . . . (a_n x +a_1)$

Then for corresponding linear factor ,

$\sf\to \red{\dfrac{A_1}{px+q},\dfrac{A_2}{(px+q)^2}, ... ..,\dfrac{A_n}{(px+q)^n} }$

Then for corresponding non-linear repeated Fractions ,

$\sf\to \red{\dfrac{B_1}{a_1x+a_1},\dfrac{B_2}{a_2x+a_2}, ... ..,\dfrac{B_n}{a_nx+a_n} }$

For example of this type refer to the attachment.

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T Y P E - 3 :-

When g(x) contains non - repeated non - reducible factors in form of a quadratic equation. that is px² + qx + c . The Partial Fraction for this type is ,

$\sf\to\red{ \dfrac{Ax + B }{px^2+qx+c}}$

where p , q , A and B belong to Real Numbers .