Question

the integral of 1/(x^2+9)

Answer 1

I cannot use ln, no?

Answer 2

You can do a trig substitution:
\(x=3\tan\theta\)
and by this, you get:
\(dx=3\sec^2\theta ~ d\theta\)

Answer 3

oh oppsies

Answer 4

No ln's

Answer 5

oh cool. thanks. so it doesn't have to be under a radical for me to use that trig sub?

Answer 6

You can use trig sub whenever, well, if it doesn't draw you into the world of a very messy notebook

Answer 7

Yes, not necessarily in the root or anything of this sort...

Answer 8

ok thanks!! that makes it easy

Answer 9

Just that trig substitution is a general get-away from thing such as

\(\large\color{black}{ \displaystyle \int \frac{1}{\sqrt{x^2+a^2}} dx }\)

Answer 10

but, that certainly doesn't mean you can't use it in such a case....

you can "relize the derivative" or you can do the integration with trig sub... either way

Answer 11

realize (not relize)

Answer 12

i see! thanks so much!

Answer 13

Yes, no prob, just one quick post:

Answer 14

\(\large\color{black}{ \displaystyle \int \frac{1}{x^2+a^2} dx }\)

\(\large\color{black}{ \displaystyle x=a\tan\theta }\)
\(\large\color{black}{ \displaystyle dx=a\sec^2\theta{~~}d\theta }\)

\(\large\color{black}{ \displaystyle \int \frac{1}{\left(a\tan\theta\right)^2+a^2} \left( a\sec^2\theta{~~}d\theta \right) }\)

\(\large\color{black}{ \displaystyle \int \frac{a\sec^2\theta}{a^2\tan^2\theta+a^2} {~~}d\theta }\)

\(\large\color{black}{ \displaystyle \int \frac{a\sec^2\theta}{a^2\left(\tan^2\theta+1\right)} {~~}d\theta }\)

\(\large\color{black}{ \displaystyle \int \frac{a\sec^2\theta}{a^2\left(\sec^2\theta\right)} {~~}d\theta }\)

\(\large\color{black}{ \displaystyle \int \frac{1}{a} {~~}d\theta =\frac{1}{a}\theta }\)

\(\large\color{black}{ \displaystyle x=a\tan\theta }\) ---> \(\large\color{black}{ \displaystyle x/a=\tan\theta }\) ---> \(\large\color{black}{ \displaystyle \theta=\tan^{-1}\left(\frac{x}{a}\right) }\)

\(\large\color{black}{ \displaystyle \int \frac{1}{a} {~~}d\theta =\frac{1}{a}\theta=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) }\)

Answer 15

thus,

\(\large\color{black}{ \displaystyle \int \frac{1}{x^2+a^2} dx =\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) }\)

Answer 16

well, +C, but that doesn't matter for what I showed.

Answer 17

it took more time for me because I tried to make sure no typos are there... have fun:)

Answer 18

thanks! i am working it out right now i will compare it to yours when I'm done!

Answer 19

Ok:)

Answer 20

yw