Question

This is a tutorial

Show using the cover-up method that

\[
\frac{1}{(x-1)^2 \left(x^2+1\right)
(x+1)}=\frac{x+1}{4
\left(x^2+1\right)}-\frac{3}{8
(x-1)}+\frac{1}{4
(x-1)^2}+\frac{1}{8 (x+1)}
\]

Answer 1

\[\frac{1}{(x-1)^2 \left(x^2+1\right)
(x+1)}=\frac{A x+B}{
\left(x^2+1\right)}+\frac{C}{
(x-1)}+\frac{D}{
(x-1)^2}+\frac{E}{ (x+1)}
\]

Let us find D

Multiply the two sides by \((x-1)^2\) and let x=1

we get

\[
\frac{1}{4}=D
\]

Multiply the two sides by \(x+1\) and let x=-1

We get

\[
\frac{1}{8}=E
\]


Multiply the two sides by \(x^2+1\) and let x=i

We get

\[
\frac{1}{(i-1)^2 (i+1)}= A i + B\\
\frac{1}{ }+\frac{i}{4} = A i + B\\
A=B=\frac 1 4

\]

Let x=0,

we get

\[
1= B -C+ D +E\\
C= B+D+E-1=\frac 1 4+ \frac 1 4 +\frac 1 8-1=-\frac 3 8
\]

Answer 2

Small misprint for finding A and B
\[
\frac{1}{(i-1)^2 (i+1)}= A i + B\\
\frac{1}{4 }+\frac{i}{4} = A i + B\\
A=B=\frac 1 4
\]

Answer 3

very nice point letting x=i ; i've never seen it
thank u sir...

Answer 4

informative and well-written, great work @eliassaab

Answer 5

Thank you guys.

Answer 6

As a practice show using the cover-up method that
\[
\frac{100}{(x-1) (x+1)^2
\left(x^2+4\right)}=\frac{4x -4}
{x^2+4}-\frac{9}{x+1}-\frac{10}{(x+1)^2}+\frac{5}{x-1}

\]

Answer 7

gr8 tutorial thanks