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)
I think that but you need x^2+ 16 into a square root to use that method.
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:
\(\displaystyle \displaystyle{\int \frac{dx}{x^2+16}}\)
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.
I fixed it above.
Wolfram alpha result is same - except they simplified further the sin(2Θ) term.