Question

How do you find `\int \frac { x } { ( 1+ 4x ^ { 2} ) ^ { 2} } d x`?

Answer 1

The answer is `=-1/(8(1+4x^2))+C`

Explanation:

We perform this integral by substitution

Let `u=1+4x^2`

`du=8xdx`

`xdx=1/8du`

So,

`int(xdx)/(1+4x^2)^2`

`=1/8int(du)/(u^2)`

`=-1/8u^-1`

`=-1/(8(1+4x^2))+C`

Answer 2

`-1/(8+32x^2)+C`

Explanation:

We define a new variable `u=1+4x^2`. Now, `(du)/dx=8x`. We then divide both sides by `8x` and multiply both sides by `dx` (an abuse of notation) to get `dx=(du)/(8x)`.

Now, let's look at our original integral: `int\ x/(1+4x^2)^2\ dx`. We can substitute the above variables to get `int\ x/(8xu^2)\ du=int\ 1/(8u^2)\ du`. Remember that you can always factor a constant out of the integral: `1/8int\ 1/(u^2)\ du`. We know that `1/u^2` is simply equal to `u^-2`: `1/8int\ u^-2\ du`. We apply the rule that `int\ x^a\ dx=x^(a+1)/(a+1)+C` for a constant `a` on this integral to get `1/8(u^(-2+1)/(-2+1)+C)=1/8(u^-1/-1+C)=-1/(8u)+1/8C`.

Since `C` is just an arbitrary constant, `1/8C` is also just an arbitrary constant. We can just replace `1/8C` by `C` to get `1/8(u^-1/-1+C)=-1/(8u)+C`.

However, we want our answer to be in terms of `x`. We said before that `u=1+4x^2`. We just need to substitute this in to get our final answer of `-1/(8+32x^2)+C`.