## anonymous 4 years ago find the following indefinite integral using u-substitution to solve... HELP

1. anonymous

$\int\limits_{}^{}6x \sqrt{x^2-25}dx$

2. anonymous

I know I let u=x^2-25 which makes du=2xdx

3. anonymous

i am stuck at a particular step.. i will show you what i have so far

4. mathmate

hint: try u=sqrt(x^2-25) or try differentiating sqrt(x^2-25) and see what you get.

5. anonymous

ok then dx = du/2x $\int\limits_{}^{}\frac{6x \sqrt{u}}{2x} du =3 \int\limits_{}^{}\sqrt{u} du$

6. anonymous

$\int\limits_{}^{}\sqrt{x^2-25}\times6xdx = 3\int\limits_{}^{}\sqrt{x^2-25}\times2xdx = 3\int\limits_{}^{}\sqrt{u}du$

7. anonymous

i'm lost after that

8. anonymous

$\sqrt{u} = u^{1/2}$ use the power rule

9. anonymous

$\large \int\limits_{}^{}x^{n} = \frac{x^{n+1}}{n+1}$

10. anonymous

yeah so i would fill fill in x^2 -25 now for that step or no?

11. anonymous

no leave it in terms of u until you have finished integrating, then the last step will be to substitute the x^2 -25 back in for u

12. anonymous

ok so i got 2u^(3/2)+c ... is that correct?

13. anonymous

yep

14. anonymous

ok so then substitute in now? giving me.. 2(x^2-25)^(3/2) + c ..?

15. anonymous

yes you could also write it as ...2sqrt(x^2-25)(x^2 -25) + c

16. anonymous

ok so then my final answer is $2x^2-50\sqrt{x^2-25}+c$..? or just leave it $2(x^2-25)\sqrt{x^2-25}+c$?

17. anonymous

i would just leave it in factored form, but both are correct

18. anonymous

thank you so much!