Evaluate the integral by substitution

[rragas]

(long S) (dx) / (x^2 +9)

u= (x/3)

i dont see how i can substitue that..

im gonna die.. this is supposed to be an easier question.. if someone can help me out today 1-on-1 on a messenger.. it would be greatly appreciated

Please let me know if anyone can.

 

[galactus]

You should recognize this as a variation on \(\displaystyle \L\\\int\frac{1}{x^{2}+1}dx=tan^{-1}(x)\)

Let's do a general case:

\(\displaystyle \L\\\int\frac{1}{x^{2}+a^{2}}dx\)

Let \(\displaystyle x=au \;\ and \;\ dx=adu\)

Then:

\(\displaystyle \L\\\int\frac{1}{a^{2}+x^{2}}dx=\int\frac{a}{a^{2}+a^{2}u^{2}}du=\frac{1}{a}\int\frac{1}{1+u^{2}}du=\frac{1}{a}tan^{-1}(u)+C=\frac{1}{a}tan^{-1}(\frac{x}{a})+C\)

Sub in what you need.

 

[rragas]

Aha.. I see.

But i still don't understand how to do it.

 

[galactus]

I edited. Check again. Does that help?.

 

[rragas]

I think I got it, thanks.