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## anonymous 4 years ago Integrate using partial fractions (x-1)/(x^2(x^2+1))dx

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1. anonymous

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

2. anonymous

$A(x^2+1)+Bx(x^2+1)+(Cx+D)x^2=x-1$

3. anonymous

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

4. anonymous

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

5. anonymous

yes thank you!

6. anonymous

$\frac{x-1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1}$ Multilpy both sides by x^2 + 1 and make x = i$\frac {i-1} {-1} = Ci + D= -i+1\\ C=-1\\ D=1\\$

7. anonymous

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

8. anonymous

thank you sir

9. anonymous

yw

10. anonymous

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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