Question

F(x)=∫1/t^2+9

Answer 1

recall that the integral of 1/(t^+1) = inverse tan of t

Answer 2

\[\int\limits \frac{ 1 }{ 1+t^2 } = \tan ^{-1}t\]

Answer 3

need to make the denominator look like that

Answer 4

let the denominator go to this ---

Answer 5

\[\int\limits \frac{ 1 }{ 9[\frac{x^2 }{ 9 } + 1] }\]

Answer 6

=
\[\frac{ 1 }{ 9 } \int\limits \frac{ 1 }{ (\frac{ x }{ 3 })^2 + 1 }\]

Answer 7

let u = x/3
du = (1/3)dx or dx = 3du

Answer 8

\[\frac{ 1 }{ 9 } \int\limits \frac{ 1 }{ u^2 + 1 } (3* du)\]

Answer 9

\[\frac{ 1 }{ 3 } \tan ^{-1}(u)\]

Answer 10

sub back in u = x/3

Answer 11

\[\frac{ 1 }{ 3 }\tan ^{-1}(\frac{ x }{ 3 })\]

Answer 12

get it all?

Answer 13

why did you remove x/3 for u?

Answer 14

well more like how. I understand why so you can make the equation equal to arctan, but how can you just randomly replace u for x/3

Answer 15

It is a substitution, If is ok if you change the integration to be over du, which is 1/3 dx