Mathematics

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

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

##### SOLUTION
Let $x-1=y$, i.e., $x=1+y$
$\therefore \displaystyle \frac{x^4}{(x-1)^4(x+1)}= \frac{(1+y)^4}{y^4(1+y+1)}$

$= \displaystyle \frac{1+4y+6y^2+4y^3+y^4}{y^4(2+y)}$
Divide $1+4y+6y^2+4y^3+y^4$ by $2+y$ and continue till the remaider is $y^4$.
$2+y 1+4y+6y^2+4y^3+y^4 \downharpoonright \frac{1}{2}+\frac{7}{4}y+\frac{17}{8}y^2+\frac{15}{16}y^3$
$1+\frac{1}{2}y$
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$\frac{7}{2}y+6y^2$
$\frac{7}{2}y+\frac{7}{6}y^2$
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$\frac{17}{4}y^2+4y^3$
$\frac{17}{4}y^2+\frac{17}{8}y^3$
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$\frac{15}{8}y^3+y^4$
$\frac{15}{8}y^3+\frac{15}{16}y^4$
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$\frac{1}{16}y^4$
$\therefore \displaystyle \frac{1+4y+6y^2+4y^3+y^4}{(2+y)}=\left[\frac{1}{2}+\frac{7}{4}y+\frac{17}{8}y^2+\frac{15}{16}y^3+\frac{\frac{1}{16}y^4}{2+y}\right]$

$\therefore \displaystyle \frac{(1+4y+6y^2+4y^3+y^4}{y^4(2+y)}=\frac{1}{2y^4}+\frac{7}{4y^3}+\frac{17}{8y^2}+\frac{15}{16y}+\frac{1}{16}\cdot\frac{1}{2+y}$

in last putting $y=x-1$
$\therefore \displaystyle \frac{x^4}{(x-1)^4(x+1)} = \frac{1}{2(x-1)^4}+\frac{7}{4(x-1)^3}+\frac{17}{8(x-1)^2}+\frac{15}{16(x-1)}+\frac{1}{16}\cdot \frac{1}{(x+1)}$
which are the required partial fractions.

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

#### Realted Questions

Q1 Subjective Medium
Evaluate the following : $\displaystyle\int \dfrac{1}{1+x-x^{2}}.dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The integral $\int _{ 0 }^{ \pi }{ \sqrt { 1+4\sin { ^{ 2 }\frac { x }{ 2 } -4 } \sin { \frac { x }{ 2 } } } dx }$ is equal to
• A. $\frac{2\pi}{3}-4-4\sqrt{3}$
• B. $4sqrt{3}-4$
• C. $4sqrt{3}-4-\frac{\pi}{3}$
• D. $\pi-4$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle I = \int \frac {e^{3x}}{1 + e^x} dx$, then I equal
• A. $\displaystyle (1/2)(1 + e^x)^2 - (1/3)(1 + e^x) + \log (1 + e^x) + C$
• B. $\displaystyle (1/2)(1 + e^x) (e^x + 3) + \log (1 + e^x) + C$
• C. $\displaystyle (1/2)(1 + e^x)^2 - 2 \log (1 + e^x) + C$
• D. $\displaystyle (1/2)(1 + e^x)(e^x - 3) + \log (1 + e^x) + C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve $\displaystyle\int{\dfrac{-3}{{{\left( x+2 \right)}^{2}}}}dx$

Show that $\displaystyle\int \sqrt{4+8x-5x^{2}}dx=\sqrt{5}\left [ \frac{5x-4}{10\sqrt{\left ( 5 \right )}}\sqrt{4+8x-5x^{2}}+\frac{18}{25}\sin ^{-1}\frac{5x-4}{6} \right ].$