Question

Given the integral of (1/(1+9x^2)dx taking u =3x.
Use integration by substitution
Question is can someone explain to me how they got u as 3x.As for me would suggest u=1+9x^2

Answer 1

let u = 3x
du = 3dx
The integral can be rewritten as:
(1/3)*(3dx/(1+9x^2))

Answer 2

but hwy 3x

Answer 3

Now if we substitute u = 3x, du = 3dx, we get:
(1/3)*(du/(1+u^2) = (1/3) (1/(1+u^2)*du

Answer 4

Because the bottom seems to be in the form of 1+a^2, where a^2 = 9x^2 = (3x)^2

Answer 5

So after you substitute u:
(1/3) (1/(1+u^2)*du, notice that the integral is just inverse tangent

Answer 6

So you can leave the constant alone, and if you integrate (1/(1+u^2))*du, you'll get ArcTan(u)

Answer 7

Now just multiply by that constant outside to get (1/3)*ArcTan(u), but u're not done yet:
u = 3x, so now we have:
(1/3)*Arctan(3x) + C (don't forget the arbitrary constant, this is an indefinite integral).