Integrate Trig Expression: $\int_\frac{dx}{(4+x^2)^2}dx$

[Aerosion]

Homework Statement

$$\int_\frac{dx}{(4+x^2)^2}dx$$

Homework Equations

The Attempt at a Solution

SO...I started by making $$x = 2*tan x$$ and $$dx = 2*sec(x)^2$$. The x is supposed to be the 0 sign with the line through it, but I don't know how to make that.

I then made the equation $$\int \frac{2*sec(x)^2}{(4+2*tan(x)^2}*2*sec(x)^2$$. I multiplied the two secants to get $$4*sec(x)^4$$ on the top, and then I turned the $$2*tan(x)^2$$ on the bottom into $$2*sec(x)^2-2$$. The equation now looks like $$\int \frac{4*sec(x)^4}{(4+2*sec(x)^2-2}$$. How does this simplify? I want to get rid of the [/tex]2*sec(x)^2[/tex] by dividing it with the top thing, but I don't think I can do that becaause of the -2 attached to it.

 

[ChaoticLlama]

you should fix your latex formatting.

your integral becomes cos²(x).

Show your work again, and let us know if you get stuck.

 

[dextercioby]

$$\int \left(\frac{1}{-2x}\right)\left(\frac{-2x}{(4+x^2)^2}{}dx\right)=-\frac{1}{2x}\cdot\frac{1}{4+x^2}+\frac{1}{2}\int \frac{dx}{x^2 (4+x^2)}$$

$$=-\frac{1}{2x}\cdot\frac{1}{4+x^2}+\frac{1}{8}\int \left(\frac{1}{x^2}-\frac{1}{4+x^2}\right){}dx$$

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