Question

integral of 5/(9+4r^2) dr

Answer 1

Before suggesting anything, what methods do you know for integration?

Answer 2

u-sub
hmm i am aware of trig and partial fractions but our teacher hasn't taught

Answer 3

i just get to 5 int (1/(9+r^2)) then i am stuck, i can't find a way to u-sub or "add zero" to make the fraction better

Answer 4

\[\LARGE\color{blue}{ \int\limits_{ }^{ } \frac{5}{4r^2+9}~~dr}\]
\[\LARGE\color{blue}{ \frac{5}{9}\int\limits_{ }^{ } \frac{1}{\frac{4r^2}{9}+1}~~dr}\]
and then substitute `u` for `2r/3`

Answer 5

I'll wait for Solomon to reply, he might have it~

Answer 6

I am not very good at this though, because the heights math I took was trig.

Answer 7

I can finish if you want to .

Answer 8

the highest...

Answer 9

yes please
But what made you decide to divide the denominator by 9?

Answer 10

I am taking the constant out, which is 5/9 in this case.

Answer 11

I don't want to take the responsibility..... @luigi, please -:( :)

Answer 12

hmm using 2r/3 for you i get 15/18 sintegral of 1/(1+u2)?

Answer 13

You're one the right path so far Solomon, now just use \(\Large \int \frac{1}{1+u^2}~du=tan^{-1}u+C\)

Answer 14

AH!
yes haha i forgot about the inverse function!

Answer 15

I was doing it correctly, and I am getting

\[\LARGE\color{blue}{ \frac{5}{9}\int\limits_{ }^{ } \frac{1}{\frac{4r^2}{9}+1}~~dr}\]