# Help me with this integral

1. May 19, 2006

### ndnbolla

Here is the integral I am suppose to solve for.

int(16/((x^2)(x^2+4))

I am using the method of integration by parts.

Let u = 16/x^2
dv = 1/(x^2+4) = 1/(x^2 + 2^2)

Then du = -32/x^3 and v = (1/2)arctan(x/2) ... by use of integral table

So, then with u*v- int(v*du) ... (16/x^2)*((1/2)arctan(x/2)) - int((1/2)arctan(x/2)*(-32/x^3) dx

Now I solved the int((1/2)arctan(x/2)*(-32/x^3) dx
Let u= (1/2)arctan(x/2)
dv = -32/x^3

Then du = 1/(x^2 + 2^2) and v=16/x^2

So I go on to solve this integral (1/2)arctan(x/2)*(16/x^2) - int(16/((x^2)(x^2+4))

The problem is when I add like terms to solve the whole integral, the term (1/2)arctan(x/2)*(16/x^2) cancels out and here is where I get stuck. I know that I am doing something wront with the above math but I just can't seem to find it.

Their is a lower limit of 2 and an upper limit of 4 but that is irrelevant for now.

2. May 19, 2006

### benorin

What you did was integrate by parts twice: the second application of it undid the first, hence you are left with the integral you started with. Use instead partial fraction decomposition, namely use the fact that

$$\frac{16}{x^2(x^2+4)}=\frac{4}{x^2}-\frac{4}{x^2(x^2+4)}$$

3. May 19, 2006

### VietDao29

Just a little typo here, benorin, the LaTeX line should read:
$$\frac{16}{x ^ 2 (x ^ 2 + 4)} = \frac{4}{x ^ 2} - \frac{4}{x ^ 2 + 4}$$
:)

4. May 19, 2006

### Hammie

You are using integration of parts twice here, once in one direction- using 1/x as u to get du of -1/x^3, and then using -1/x^3 as dv, finding 1/x^2 as v. Essentially, you are going in circles.

You may find partial fraction decomposition an easier method prior to trying to integrate.

5. May 21, 2006

### benorin

Thank you VietDao29.