## anonymous one year ago Compute the integral.

1. anonymous

2. tacotime

You can change the exponents to negatives to move the denominator to the numerator.

3. freckles

partial fractions should work

4. freckles

$\frac{1}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}$

5. anonymous

Yeah i got that , having difficulties finding A,B,C, and D

6. freckles

which method are you used to: heaviside or comparing and solving a system of linear equations?

7. freckles

$Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1$

8. anonymous

both were taught i suppose. i dont mind expanding everything

9. freckles

for example heaviside is pluggin in numbers to help solve like replace x with 0 and we get B(0+4)=1 so B=1/4 then you can plug in 2i for x $(C \cdot 2i+D)(4i^2)=1 \\ (C \cdot 2i+D)(-4)=1 \\ C \cdot 2i+D=\frac{-1}{4}$ plug in -2i for x $(C \cdot (-2i)+D)(4i^2)=1 \\ -2i C+D=\frac{-1}{4} \\ 2iC+D=\frac{-1}{4} \text{ which was given by the other x value we plugged \in } \\ \text{ now add } 2D=\frac{-2}{4}=\frac{-1}{2} \\ \text{ so } D=\frac{-1}{4} \\ \text{ so now } C \\ -2i C-\frac{1}{4}=\frac{-1}{4} \\ -2i C=0 \\ \text{ so we have } C=0$

10. freckles

but you probably don't like imaginary numbers... we could just do the whole expanding thing then comparing then solving the system thing we ended up solving a system for the heaviside anyways

11. anonymous

it's definately not the heaviside lol

12. freckles

I only plugged in -2i and 2i because it made evaluating x^2+4 simple

13. freckles

since (2i)^2+4 is 0 and (-2i)^2+4 is also 0

14. freckles

ok you expand and I will expand and regroup terms and you can compare my equation to your equation

15. freckles

$Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \\ A(x^3+4x)+B(x^2+4)+Cx^3+Dx^2 =1 \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=1$

16. anonymous

|dw:1444155517948:dw|

17. freckles

great we have no x^3 's on the other side so you should see A+C=0 we have no x^2 's on the other side so you should see B+D=0 we have no x 's on the other side so you should see 4A=0 we do have constants on both sides so we need 4B=1

18. freckles

I would solve the last two equations first because they only involve one constant variable each

19. freckles

$A=0 , B=\frac{1}{4}$

20. freckles

and then go to the top two equations and used these values to solve for C and D

21. anonymous

hold on, i always get stuck on this part. you get A+C=0 .B+D=0. 4A=0. 4B=1 by comparing the variable son each side?

22. freckles

yeah yeah

23. freckles

$Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \\ A(x^3+4x)+B(x^2+4)+Cx^3+Dx^2 =1 \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=1 \text{ same as } \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=x^3(0)+x^2(0)+x(0)+1$

24. freckles

we want both sides to be the same thing this can only happen if A+C and B+D and 4A are zero and 4B is one.

25. anonymous

A=0 B=1/4 C=0 D=-1/4

26. freckles

sounds awesome so you have: $\int\limits \frac{1}{4} \frac{1}{x^2} dx-\int\limits \frac{1}{4} \cdot \frac{1}{x^2+4} dx$

27. freckles

$\frac{1}{4} \int\limits x^{-2} dx-\frac{1}{4} \int\limits \frac{1}{x^2+4} dx \\ \frac{1}{4} \int\limits x^{-2} dx - \frac{1}{4 } \int\limits \frac{\frac{1}{4}}{\frac{x^2}{4}+1} dx \\ \frac{1}{4} \int\limits x^{-2} dx-\frac{1}{4} \cdot \frac{1}{4} \int\limits \frac{1}{(\frac{x}{2})^2+1} dx \\ \frac{1}{4 } \int\limits x^{-2} -\frac{1}{16} \int\limits \frac{1}{(\frac{x}{2})^2+1} dx$

28. anonymous

coool. i think is got it from here. i should be able to integrate . i was just confused on the decomposition part. thank you!

29. freckles

np

30. anonymous

i got it from here*