Question

How do you evaluate the integral `int x^2/(4+x^2)^2dx`?

Answer 1

`intcolor(white)(l)(x^2)/((4+x^2)^2)color(white)(l)dx = color(blue)(1/4(tan^-1[x/2] - (2x)/(x^2+4)) + C`

Explanation:

We're asked to find the integral

`intcolor(white)(l)(x^2)/((x^2+4)^2)color(white)(l)dx`

Let's split this up into partial fractions:

`= intcolor(white)(l)1/(x^2+4) - 4/((x^2+4)^2)color(white)(l)dx`

`= intcolor(white)(l)1/(x^2+4)color(white)(l)dx - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx`

`= intcolor(white)(l)1/(4((x^2)/4+1))color(white)(l)dx - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx`

`= 1/4intcolor(white)(l)1/(4((x^2)/4+1))color(white)(l)dx - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx`

For the integrand `1/((x^2)/4+1)`, let's make `u = x/2` and `dx = 2color(white)(l)du`:

`= 1/2intcolor(white)(l)1/(u^2+1)color(white)(l)du - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx`

The integral of `1/(u^2+1)` is `tan^-1u`:

`= 1/2tan^-1u - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx`

For the integrand `1/((x^2+4)^2)`, let's make `x = 2tans` and `dx = 2sec^2scolor(white)(l)ds`.

Then `(x^2+4)^2 = (4tan^2s+4)^2 = 16sec^4s` and `s = tan^-1[x/2]`:

`= 1/2tan^-1u - 8intcolor(white)(l)1/16cos^2scolor(white)(l)ds`

`= 1/2tan^-1u - 1/2intcolor(white)(l)cos^2scolor(white)(l)ds`

We can write `cos^2s` as `1/2cos[2s] + 1/2`:

`= 1/2tan^-1u - 1/2intcolor(white)(l)1/2cos[2s] + 1/2color(white)(l)ds`

`= 1/2tan^-1u - 1/4intcolor(white)(l)cos[2s]color(white)(l)ds - 1/4intcolor(white)(l)1color(white)(l)ds`

For the integrand `cos[2s]`, let's make `p = 2s` and `ds = 1/2dp`:

`= 1/2tan^-1u - 1/8intcolor(white)(l)cospcolor(white)(l)dp - 1/4intcolor(white)(l)1color(white)(l)ds`

The integral of `cosp` is `sinp`:

`= 1/2tan^-1u - 1/8sinp - 1/4intcolor(white)(l)1color(white)(l)ds`

The integral of `1` is `s`:

`= 1/2tan^-1u - 1/8sinp - 1/4s + C`

Substitute back in `p = 2s`:

`= 1/2tan^-1u - 1/8sin[2s] - 1/4s + C`

Substitute back in `s = tan^-1[x/2]`:

`= ((x^2+4)(2tan^-1u - tan^-1[x/2]) - 2x)/(4(x^2+4)) + C`

Substitute back in `u = x/2`:

`= ((x^2+4)tan^-1[x/2] - 2x)/(4(x^2+4)) + C`

Or

`color(blue)(ulbar(|stackrel(" ")(" "I = 1/4(tan^-1[x/2] - (2x)/(x^2+4)) + C" ")|)`