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

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

Mathematics
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\[ \frac{x-1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1} \]
\[ A(x^2+1)+Bx(x^2+1)+(Cx+D)x^2=x-1 \]
Multiply bith sides by x^2 and make x =0, you get A= -1

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Do you know how to take it from there, @Silenthill ?
yes thank you!
\[ \frac{x-1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1} \] Multilpy both sides by x^2 + 1 and make x = i\[ \frac {i-1} {-1} = Ci + D= -i+1\\ C=-1\\ D=1\\ \]
\[ \frac{x-1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1} \] Multiply both sides by x and let x goes to Infinity, you get 0 = B + C B=-C=1 Putting everything together, you get \[ \frac{x-1}{x^2 \left(x^2+1\right)}=\frac{1-x}{x^2+1}-\frac{1}{x^2}+\frac{1}{x} \]
thank you sir
yw
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

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