## anonymous 5 years ago need help with evaluating this integral.... (x^9)/(sqrt(3+x^5)) dx.....

1. anonymous

I would recommend a u substitution of u = x^5+3

2. anonymous

ok i started that....

3. anonymous

How far did you get?

4. anonymous

then i got (1/5)du = x^4dx ....

5. anonymous

Ok, though I prefer dx = du/(5x^4)

6. anonymous

ok.... then

7. anonymous

So then, what is x^5 in terms of u?

8. anonymous

i dont know what you mean by that

9. anonymous

u = x^5 + 3 so x^5 = ?

10. anonymous

x^5 = u - 3 right?

11. anonymous

u - 3

12. anonymous

ok. Good. Now start substituting in your integral

13. anonymous

so whats the x^9?... the denominator will be sqrt3?

14. anonymous

The denominator will be $$\sqrt{u}$$

15. anonymous

thats what i meant.///

16. anonymous

the dx will be $\frac{1}{5x^4}du$

17. anonymous

So what's left in the numerator( after you simplify)

18. anonymous

idk.......

19. anonymous

Cmon. $\int \frac{x^9}{\sqrt{x^5+3}}dx = \int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx$ $\text{Let }u = x^5+3$$\implies x^5 = u-3$ $\implies 5x^4dx = du \implies dx = \frac{1}{5x^4}du$ Therefore $\int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx = \int \frac{(u-3)*x^4}{\sqrt{u}} (\frac{1}{5x^4})du$

20. anonymous

And we can cancel the x^4 from top and bottom. Then split the fraction into two different fractions $\frac{u}{\sqrt{u}} - \frac{3}{\sqrt{u}}$

21. anonymous

and integrate each one separately.

22. anonymous

what happened to the 5 in the faraction

23. anonymous

yes there should be a factor of 5 outside the integral , he most likely didnt mention it because any integral of a constant multiple is just the integral times that constant