Question

integral of - infinity to infinity of 1/(x^2+4)

Answer 1

note that \(\frac{1}{x^2+1}=\frac{1}{4}\frac{1}{(\frac{x}{2})^2+1}\)
so
\(\int\frac{1}{x^2+4}dx=\frac{1}{4}\int\frac{1}{(\frac{x}{2}+1)}=2*\frac{1}{4}\tan^{-1}(\frac{x}{2})\)

Answer 2

so we can see it relates to arctan so we make it that?

Answer 3

exactly

Answer 4

the 2 on the outside is to deal with the chain rule on the x/2

take the derivative of the answer and you will see what I mean.

Answer 5

ok so if i got a general formula
integral of 1/(x^2+a^2)

Answer 6

we would facotor out 1/a

Answer 7

looks like it
\(\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\tan^{-1}(\frac{x}{a})\)

Answer 8

ok sorry but can you explain how you got x/2 when u factored our 1/4
i want to fully understand the algebra!

Answer 9

or even better
\(\int\frac{1}{x^2+b}dx=\frac{1}{\sqrt{b}}\tan^{-1}(\frac{x}{\sqrt{b}})\) when \(b>0\)

Answer 10

ahh ok

Answer 11

\(\frac{1}{x^2+4}=\frac{1}{4}\frac{1}{\frac{x^2}{4}-\frac{4}{4}}=\frac{1}{4}\frac{1}{\frac{x^2}{2^2}-1}=\frac{1}{4}\frac{1}{(\frac{x}{2})^2-1}\)

Answer 12

same deal as \(x^2+4=4(\frac{x^2}{4}+1)=4((\frac{x^2}{2})-1)\)

Answer 13

ooooohh makes sense so when we actor in 1/4 we get what we had before duhh .. wow that was really simple

Answer 14

when we factor something out we are really just multiplying by 1
\(a+b=1*(a+b)=\frac{a}{a}(a+b)=a(\frac{a}{a}+\frac{b}{a})=a(1+\frac{b}{a})\)

Answer 15

this should also make it clear why we can never "factor out" a zero.:)

Answer 16

ok yah! dude thanks this is alot of help!

Answer 17

np:)