Question

find the following indefinite integral using u-substitution to solve... HELP

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

ok then dx = du/2x

\[\int\limits_{}^{}\frac{6x \sqrt{u}}{2x} du =3 \int\limits_{}^{}\sqrt{u} du\]

Answer 6

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

Answer 7

i'm lost after that

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

yep

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

thank you so much!