What is partialfraction (a^2+a^2+a+1)/(a^4-1) ?

The solution to partialfraction (a^2+a^2+a+1)/(a^4-1) is (-a+1)/(2(a^2+1))-1/(2(a+1))+1/(a-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Algebra >

partialfraction (a^2+a^2+a+1)/(a^4-1)

Solution

partial fractions

\frac{a ^{2} + a ^{2} + a + 1}{a ^{4} - 1}

Solution

\frac{- a + 1}{2 ( a ^{2} + 1 )} - \frac{1}{2 ( a + 1 )} + \frac{1}{a - 1}

Solution steps

\frac{a ^{2} + a ^{2} + a + 1}{a ^{4} - 1}

Expand

a^{2} + a^{2} + a + 1

\frac{2 a ^{2} + a + 1}{a ^{4} - 1}

Factor

a^{4} - 1 : (a^{2} + 1) (a + 1) (a - 1)

= \frac{2 a ^{2} + a + 1}{( a + 1 ) ( a - 1 ) ( a ^{2} + 1 )}

Create the partial fraction template using the denominator

(a^{2} + 1) (a + 1) (a - 1)

\frac{2 a ^{2} + a + 1}{( a ^{2} + 1 ) ( a + 1 ) ( a - 1 )} = \frac{a _{1} a + a}{a ^{2} + 1} + \frac{a _{2}}{a + 1} + \frac{a _{3}}{a - 1}

Multiply equation by the denominator

\frac{( 2 a ^{2} + a + 1 ) ( a ^{2} + 1 ) ( a + 1 ) ( a - 1 )}{( a ^{2} + 1 ) ( a + 1 ) ( a - 1 )} = \frac{( a _{1} a + a ) ( a ^{2} + 1 ) ( a + 1 ) ( a - 1 )}{a ^{2} + 1} + \frac{a _{2} ( a ^{2} + 1 ) ( a + 1 ) ( a - 1 )}{a + 1} + \frac{a _{3} ( a ^{2} + 1 ) ( a + 1 ) ( a - 1 )}{a - 1}

Simplify

2 a^{2} + a + 1 = (a_{1} a + a) (a + 1) (a - 1) + a_{2} (a^{2} + 1) (a - 1) + a_{3} (a^{2} + 1) (a + 1)

Solve the unknown parameters by plugging the real roots of the denominator:

- 1, 1

For the denominator root

- 1 : a_{2} = - \frac{1}{2}

For the denominator root

1 : a_{3} = 1

a_{2} = - \frac{1}{2}, a_{3} = 1

Plug in the solutions to the known parameters

2 a^{2} + a + 1 = (a_{1} a + a) (a + 1) (a - 1) + (- \frac{1}{2}) (a^{2} + 1) (a - 1) + 1 \cdot (a^{2} + 1) (a + 1)

Expand

2 a^{2} + a + 1 = a_{1} a^{3} + \frac{a ^{3}}{2} + \frac{3 a ^{2}}{2} + \frac{a}{2} + \frac{1}{2} + a a^{2} - a_{1} a - a + 1

Extract Variables from within fractions

2 a^{2} + a + 1 = a_{1} a^{3} + \frac{1}{2} a^{3} + \frac{3}{2} a^{2} + \frac{1}{2} a + \frac{1}{2} + a a^{2} - a_{1} a - a + 1

Group elements according to powers of

x

2 a^{2} + 1 \cdot a + 1 = a^{3} (a_{1} + \frac{1}{2}) + a^{2} (a + \frac{3}{2}) + a (\frac{1}{2} - a_{1}) + (1 + \frac{1}{2} - a)

Equate the coefficients of similar terms on both sides to create a list of equations

\frac{1}{2} - a + 1 = 1 \frac{1}{2} - a_{1} = 1 \frac{3}{2} + a = 2 a_{1} + \frac{1}{2} = 0

Solve system of equations:

a = \frac{1}{2}, a_{1} = - \frac{1}{2}

a = \frac{1}{2}, a_{1} = - \frac{1}{2}

Plug the solutions to the partial fraction parameters to obtain the final result

\frac{( - \frac{1}{2} ) a + \frac{1}{2}}{a ^{2} + 1} + \frac{( - \frac{1}{2} )}{a + 1} + \frac{1}{a - 1}

Simplify

\frac{- a + 1}{2 ( a ^{2} + 1 )} - \frac{1}{2 ( a + 1 )} + \frac{1}{a - 1}

Popular Examples

x^2+(2x+2)^2=(3x-2)^2

x^{2} + (2 x + 2)^{2} = (3 x - 2)^{2}

7.2>0.9(n+8.6)

7.2 > 0.9 (n + 8.6)

4(-2x+1)=6-(2x-4)

4 (- 2 x + 1) = 6 - (2 x - 4)

factor a^2-28a+196

f a c t or a^{2} - 28 a + 196

simplify (3-6n^5-8n^4)-(-6n^4-3n-8n^5)

s im pl i f y (3 - 6 n^{5} - 8 n^{4}) - (- 6 n^{4} - 3 n - 8 n^{5})

Frequently Asked Questions (FAQ)

What is partialfraction (a^2+a^2+a+1)/(a^4-1) ?
The solution to partialfraction (a^2+a^2+a+1)/(a^4-1) is (-a+1)/(2(a^2+1))-1/(2(a+1))+1/(a-1)

partialfraction (a^2+a^2+a+1)/(a^4-1)

Solution

Solution

Popular Examples

Frequently Asked Questions (FAQ)

What is partialfraction (a^2+a^2+a+1)/(a^4-1) ?

Popular topics