Resolve x4-x-14x-1 into partial fractions-

The correct option is B 1/2(x 1)^4+7/4(x 1)^3+17/8(x 1)^2+15/16(x 1)+1/16·1/(x+1) Let x 1=y , i.e., x=1+y ∴ nbsp; nbsp; nbsp; nb ...

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Standard XII

Mathematics

Integration by Partial Fractions

Resolve x4/...

Question

Resolve $\frac{x^{4}}{(x - 1)^{4} (x + 1)}$ into partial fractions.

\frac{1}{2 (x - 1)^{4}} - \frac{7}{4 (x - 1)^{3}} + \frac{17}{8 (x - 1)^{2}} + \frac{15}{16 (x - 1)} + \frac{1}{16} \cdot \frac{1}{(x + 1)}

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\frac{1}{2 (x - 1)^{4}} + \frac{7}{4 (x - 1)^{3}} + \frac{17}{8 (x - 1)^{2}} + \frac{15}{16 (x - 1)} + \frac{1}{16} \cdot \frac{1}{(x + 1)}

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\frac{1}{2 (x - 1)^{4}} + \frac{5}{4 (x - 1)^{3}} + \frac{17}{8 (x - 1)^{2}} + \frac{15}{16 (x - 1)} + \frac{1}{16} \cdot \frac{1}{(x + 1)}

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\frac{1}{2 (x - 1)^{4}} + \frac{7}{4 (x - 1)^{3}} + \frac{13}{8 (x - 1)^{2}} + \frac{15}{16 (x - 1)} + \frac{1}{16} \cdot \frac{1}{(x + 1)}

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Solution

The correct option is B $\frac{1}{2 (x - 1)^{4}} + \frac{7}{4 (x - 1)^{3}} + \frac{17}{8 (x - 1)^{2}} + \frac{15}{16 (x - 1)} + \frac{1}{16} \cdot \frac{1}{(x + 1)}$
Let $x - 1 = y$ , i.e., $x = 1 + y$
$∴ \frac{x^{4}}{(x - 1)^{4} (x + 1)} = \frac{(1 + y)^{4}}{y^{4} (1 + y + 1)}$
$= \frac{1 + 4 y + 6 y^{2} + 4 y^{3} + y^{4}}{y^{4} (2 + y)}$
Divide $1 + 4 y + 6 y^{2} + 4 y^{3} + y^{4}$ by $2 + y$ and continue till the remaider is $y^{4}$ .
$2 + y 1 + 4 y + 6 y^{2} + 4 y^{3} + y^{4} ⇂ \frac{1}{2} + \frac{7}{4} y + \frac{17}{8} y^{2} + \frac{15}{16} y^{3}$
$1 + \frac{1}{2} y$
$- -$
________
$\frac{7}{2} y + 6 y^{2}$
$\frac{7}{2} y + \frac{7}{6} y^{2}$
$- -$
_
$\frac{17}{4} y^{2} + 4 y^{3}$
$\frac{17}{4} y^{2} + \frac{17}{8} y^{3}$
$- -$

$\frac{15}{8} y^{3} + y^{4}$
$\frac{15}{8} y^{3} + \frac{15}{16} y^{4}$
$- -$
_____
$\frac{1}{16} y^{4}$
$∴ \frac{1 + 4 y + 6 y^{2} + 4 y^{3} + y^{4}}{(2 + y)} = ⎡ ⎣ \frac{1}{2} + \frac{7}{4} y + \frac{17}{8} y^{2} + \frac{15}{16} y^{3} + \frac{\frac{1}{16} y^{4}}{2 + y} ⎤ ⎦$
$∴ \frac{(1 + 4 y + 6 y^{2} + 4 y^{3} + y^{4}}{y^{4} (2 + y)} = \frac{1}{2 y^{4}} + \frac{7}{4 y^{3}} + \frac{17}{8 y^{2}} + \frac{15}{16 y} + \frac{1}{16} \cdot \frac{1}{2 + y}$
in last putting $y = x - 1$
$∴ \frac{x^{4}}{(x - 1)^{4} (x + 1)} = \frac{1}{2 (x - 1)^{4}} + \frac{7}{4 (x - 1)^{3}} + \frac{17}{8 (x - 1)^{2}} + \frac{15}{16 (x - 1)} + \frac{1}{16} \cdot \frac{1}{(x + 1)}$
which are the required partial fractions.

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Similar questions

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

Q. The value of

(x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3} - \frac{1}{4} (x - 1)^{4} + \dots .

(x - 1)^{4} + 4 (x - 1)^{3} + 6 (x - 1)^{2} + 4 (x - 1) + 1 =

Q. If

(1 - y) (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + 32 x^{5}) = 1 - y^{6},

where

y \neq 1, x \neq \frac{1}{2}

, then the value of

\frac{y}{x}

Q. The quotient when

1 + x^{2} + x^{4} + x^{6} + \dots + x^{34}

is divided by

1 + x + x^{2} + x^{3} + \dots + x^{17}

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Resolve x4(x−1)4(x+1) into partial fractions.

Resolve $\frac{x^{4}}{(x - 1)^{4} (x + 1)}$ into partial fractions.