## ksaimouli 3 years ago by partial frac

1. ksaimouli

$\int\limits_{}\frac{ 2x^3 }{ x^2-4 }$

2. ksaimouli

i should do the long division right

3. ksaimouli

i got $x^2-4+\frac{ 8x }{ 2x }$

4. ksaimouli

@zepdrix

5. anonymous

You're right about doing long division first, but it doesn't look like you did it right.

6. ksaimouli

|dw:1361917090294:dw|

7. anonymous

|dw:1361917028271:dw|

8. anonymous

Right, so by long division you get $2x+\frac{8x}{x^2-4}$

9. ksaimouli

so should i make the whole function as|dw:1361917417172:dw|

10. anonymous

You can, but that doesn't mean you should. Your integral has changed from $\int\frac{2x^3}{x^2-4}dx$ to $\int\left(2x+\frac{8x}{x^2-4}\right)\;dx$ Integrate term by term, with partial fraction decomp on the second term.

11. ksaimouli

right here i could use u substitution right

12. anonymous

Or that, yes. A sub might actually shorten the work, so I'd suggest that.

13. ksaimouli

i got $x^2+4\ln(x^2-4)$

14. ksaimouli

+C

15. ksaimouli

is this right @SithsAndGiggles

16. anonymous

Yes that's right, though I would replace the parentheses with absolute value bars.

17. ksaimouli

ya i dont know how to use that in equation so put the ()

18. ksaimouli

thx

19. anonymous

You're welcome. By the way, absolute value bars are above the Enter/Return key. Hold shift and click the backslash (\) button to get (|)

20. ksaimouli

|2x+1|

21. ksaimouli

ya i got that thx