Question

Integrate using partial fractions (x-1)/(x^2(x^2+1))dx

Answer 1

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

Answer 2

\[
A(x^2+1)+Bx(x^2+1)+(Cx+D)x^2=x-1
\]

Answer 3

Multiply bith sides by x^2 and make x =0, you get
A= -1

Answer 4

Do you know how to take it from there, @Silenthill ?

Answer 5

yes thank you!

Answer 6

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

\]
Multilpy both sides by x^2 + 1 and make x = i\[

\frac {i-1} {-1} = Ci + D= -i+1\\
C=-1\\
D=1\\
\]

Answer 7

\[
\frac{x-1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1}
\]
Multiply both sides by x and let x goes to Infinity, you get
0 = B + C
B=-C=1
Putting everything together, you get

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

Answer 8

thank you sir

Answer 9

yw

Answer 10

given question equals

x/(x^2(x^2+1) -1/(x^2(x^2+1)
first part can be done usin u=x^2 and then applying partial factor method whereas the second part can be seperated as1/(x^2)-1/(x^2+1)
and then both the parts can be integrated easily