Calculus integration problem

Hello everyone,
I was doing the following problem: ∫ (1) / (9 + 16x^2) dx
I had to use inverse trigonometric functions to do the following integral. These are my steps:
∫ (1) / (16(9/16 + x^2)) dx <---I factor the denominator by taking out 16 here
(1/16) * ∫ (1) / ((3/4)^2 + x^2) <--- Here I took out 1/16 which left me with an inverse trig formula
Then I did the following
(1/16) * arctan(4x/3) + c <--- I got this as a final answer.
But the problem is I have the solution to this question and it has the following answer:(1/12)arctan(4x/3)+c
Could someone tell me where I went wrong and if you could please show me how to get (1/12)arctan(4x/3)+c.

Thanks in advance

Hello

∫ (1) / (9 + 16x^2) dx

"∫ (1) / (16(9/16 + x^2)) dx <---I factor the denominator by taking out 16 here "
CORRECT

"(1/16) * ∫ (1) / ((3/4)^2 + x^2) <--- Here I took out 1/16 which left me with an inverse trig formula"
CORRECT

"Then I did the following (1/16) * arctan(4x/3) + c <--- I got this as a final answer."
FALSE OBVIOUSLY

True, ∫ 1/(1 + x²).dx = arctan(x)
But.
∫ 1/(1 + (4x/3)²).dx ≠ arctan(4x/3)

Because you forgot this derivation rule:
(f○g)' = g' × f'○g

Hence, if f=arctan(x) and g=4x/3
f'=1/(1 + x²) and g'=4/3 !!!!!!!!!!!!
[arctan(4x/3)]' = 4/3 × 1/(1 + (4x/3)²)

So the correct solving of your integration was:

∫ [1/(9 + 16x²)].dx
= ∫ [(1/9) / (1 + 16x²/9)].dx
= (1/9).∫ {1/[1 + (4x/3)² }.dx
= (1/9)×(3/4).∫ {(4/3)/[1 + (4x/3)² }.dx →→ Here we add the 4/3 factor we need
= (1/12).∫ {(4/3)/[1 + (4x/3)² }.dx →→ Which obviously means, this factor must be balanced elsewhere!!!
= (1/12).arctan(4x/3) + C

Didactically,
Dragon.Jade :-)

Trig substitution is indeed needed here. You want x^2 to equal (9/16)(tanθ)^2 so you can simplify the denominator, so let...