Question

integral of 1/(16x^2 -1)

Answer 1

Lets rewrite it:\[\int\limits_{}^{}\frac{ 1 }{ (4x)^2-1 }\]
Does that look like something familiar? ;)

Answer 2

That's what i tried but what integral trig formula does that fi into I couldn't find one

Answer 3

\[\int\limits_{}^{}\frac{ dx }{ 16x ^{2} - 1 } = \int\limits_{}^{}\frac{ Adx }{ 4x - 1 } + \int\limits_{}^{}\frac{ Bdx }{ 4x + 1 }\]This technique is called partial fraction decomposition and you then solve for "A" and "B". You then find the integrals using natural logs, not trig functions.

Answer 4

4A + 4B = 0 and A - B = 1

So, A = 1/2 and B = -1/2

This is "factoring in reverse". Once you plug those values in for "A" and "B", you would be able to get back to the original integral. So now, just solve the 2 integrals on the right side.

Answer 5

\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ dx }{ 4x - 1 } - \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ dx }{ 4x + 1 }\]

Answer 6

Thank you so much!

Answer 7

uw!

Answer 8

Good luck to you in all of your studies and thx for the recognition! @zim41594

Answer 9

Can you do the rest on your own, @zim41594 or do you still need some help?

Answer 10

@tcarroll010 i kind of wanted help on this one question\[\int\limits_{}^{} \frac{ 2+x }{ \sqrt{1-x^2} } dx\]