Question

Integrate the following

Answer 1

\[\int\limits_{?}^{?}\frac{ dx }{ 16x ^{2}+1 }\]

Answer 2

I believe you want to use arctan here

Answer 3

\[\large\rm \int\limits\frac{dx}{16x^2+1}\quad=\quad \int\limits\frac{dx}{(4x)^2+1}\]This is a helpful step, bringing the 16 into the square.
Then apply a trig sub, ya? :)

Answer 4

\(\large\rm 4x=\tan\theta\)

Answer 5

ah thanks guys
im taking calc 2 but its been ages since i've done calc

Answer 6

you can use a substitution as shown, or take out a factor of x^2

\[\int\limits \frac{ dx }{ 16x^2+1 } = \frac{ 1 }{ 16 }\int\limits \frac{ dx }{ x^2 + \frac{ 1 }{ 16 } } =\frac{ 1 }{ 16 }\int\limits \frac{ dx }{ x^2 + (\frac{ 1 }{ 4 })^2 }\]
\[\frac{ 1 }{ 4 } \tan^{-1} (\frac{ x }{ \frac{ 1 }{ 4 } }) + C = \frac{ 1 }{ 4 }\tan^{-1} (4x) + C\]