Question

int(1/(x^2+4)^2dx

Answer 1

the denom is to the 2nd power

Answer 2

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

Answer 3

lol. my bad. read too fast

Answer 4

dam u wrote a lot too before u deleted that

Answer 5

you have your integral of the form\[\int\limits_{}^{}\frac{ dx }{ (x^2 + a^2)^2 }\]
to solve, you must use trigonometric substitution. for this case:
let x = atan(u). for our case. a = 2

x = 2 tan(u)
dx = 2 sec^2 (u) du

\[\int\limits_{}^{}\frac{ \sec^2(u)du }{ (4\tan^2(u) + 4)^2 } = \int\limits_{}^{}\frac{ \sec^2(u) du }{ (4(\tan^2 + 1))^2 } = \frac{ 1 }{ 16 }\int\limits_{}^{}\frac{ \sec^2(u) du }{ \sec^4(u) } \]
last step using trig id: tan^2(x) + 1 = sec^2(x)\[\frac{ 1 }{ 16 }\int\limits_{}^{} \cos^2(u) du = \frac{ 1 }{ 16 }\int\limits_{}^{}\frac{ 1 + \cos(2u) }{ 2 }du = \frac{ 1 }{ 32 }\left( u + \frac{ \sin(2u) }{ 2 } \right) + C\]

x = 2 tan(u). so:\[u = \tan^{-1}\left( \frac{ x }{ 2 } \right)\]plug u back in:
\[\frac{ 1 }{ 32 }\left( \tan^{-1}\left( \frac{ x }{ 2 } \right) + \frac{ 1 }{ 2 }\sin \left( 2 \tan^{-1}\left( \frac{ x }{ 2 } \right) \right)\right)+C\]

sin(tan^{-1}(x)) has a simplified trig value , but left as is is fine

Answer 6

wholly smokes, I sure hope this is rights lol. THank you

Answer 7

multiply everything by 2

Answer 8

i plugged in sec^2(u) du instead of 2 sec^2(u) du

makes final answer 1/16 lol

Answer 9

Basically what I meant was that I hoped he was right so he didnt have to spend anymore time making tedious corrections to the problems. Typing in the symbols is annoying and he was kind enough to put in the time to do it. I did not mean it to be rude.

Answer 10

@Euler271 why is it tan(u) and not \[\tan (\Theta)\]

Answer 11

the variable doesn't matter.

i guess i should have chosen theta because it's how i learned it at school but its really the same thing.

only by simple convenience and habit does the greek letter theta represent angle