need help with an integral

pipe

New member
Joined
Oct 31, 2014
Messages
8
hello guys i'm new, i just need help to calculate the integral: 1/(x^2 +16)^2 dx
i don't know whitch metod i have to use, i try parts, sustitution, partial fractions.
 
hello guys i'm new, i just need help to calculate the integral: 1/(x^2 +16)^2 dx
i don't know whitch metod i have to use, i try parts, sustitution, partial fractions.

Try:

x = 4 * tan(Θ)
 
hello guys i'm new, i just need help to calculate the integral: 1/(x^2 +16)^2 dx
i don't know whitch metod i have to use, i try parts, sustitution, partial fractions.

Since almost a week has passed - camera-ready solution:

\(\displaystyle \displaystyle{\int \frac{dx}{(x^2+16)^2}}\) ...... Corrected

substitute

x = 4tan(Θ)

dx = 4sec2(Θ) dΘ

\(\displaystyle \displaystyle{\int \frac{dx}{(x^2+16)^2}}\)

\(\displaystyle = \ \displaystyle{\int \frac{4sec^2(\theta) d\theta}{(16tan^2(\theta)+16)^2}}\)

\(\displaystyle = \ \displaystyle{\int \frac{4sec^2(\theta) d\theta}{256sec^4(\theta)}}\)

\(\displaystyle = \ \displaystyle{\frac{1}{64}\int cos^2(\theta)d\theta}\)

\(\displaystyle = \ \displaystyle{\frac{1}{128}\int \left [1 + cos(2\theta)\right ]d\theta}\)

\(\displaystyle = \ \displaystyle{\frac{1}{128} (\theta) + \frac{1}{256}\left [sin(2\theta)\right ] + C}\)

\(\displaystyle = \ \displaystyle{\frac{1}{128}\left [tan^{-1}(\frac{x}{4})\right ] + \frac{1}{256}\left [sin[2tan^{-1}(\frac{x}{4})]\right ] + C}\)
 
Last edited by a moderator:
Since almost a week has passed - camera-ready solution:

\(\displaystyle \displaystyle{\int \frac{dx}{x^2+16}}\)

substitute

x = 4tan(Θ)

dx = 4sec2(Θ) dΘ

\(\displaystyle \displaystyle{\int \frac{dx}{x^2+16}}\)

\(\displaystyle = \ \displaystyle{\int \frac{4sec^2(\theta) d\theta}{16tan^2(\theta)+16}}\)

\(\displaystyle = \ \displaystyle{\int \frac{4sec^2(\theta) d\theta}{16sec^2(\theta)}}\)

\(\displaystyle = \ \displaystyle{\frac{1}{4}\int d\theta}\)

\(\displaystyle = \ \displaystyle{\frac{1}{4}\theta \ + \ C}\)

\(\displaystyle = \ \displaystyle{\frac{1}{4}tan^{-1}\left (\frac{x}{4}\right ) \ + \ C}\)

you make a little mistake the integral is \(\displaystyle \displaystyle{\int \frac{dx}{(x^2+16)^2}}\)
jajajajajaja i got an answer days ago but it's complete diferent to the wolfram result.
 
you make a little mistake the integral is \(\displaystyle \displaystyle{\int \frac{dx}{(x^2+16)^2}}\)
jajajajajaja i got an answer days ago but it's complete diferent to the wolfram result.

I fixed it above.

Wolfram alpha result is same - except they simplified further the sin(2Θ) term.
 
Top