# Minus sign before the 2nd integral

1. Dec 6, 2007

### danni7070

[Solved] Where did I go wrong?

$$\int \frac{(x-2)}{x(1+x^2)}dx$$

$$\int \frac{x}{x+x^3}dx - \int \frac{2}{x+x^3}dx$$

I choose $$u = arctan(x)$$ so $$du = \frac{1}{1+x^2}$$

and $$\int \frac{x}{x+x^3} = \int \frac{1}{1+x^2}$$

1)

$$\int \frac{1}{1+x^2}dx = arctan(x)$$

2)

$$\int \frac{2}{x+x^3} = \ln (\frac{x^2}{x^2+1})$$

So $$\int \frac{(x-2)}{x(1+x^2)}dx = arctan(x) + \ln (\frac{x^2}{x^2+1})$$

Maybe I'm doing something terribly wrong but maybe not. The answer is according to Texas Instrument calculator:

$$\ln (\frac{x^2+1}{x^2}) + arctan(x)$$

Last edited: Dec 6, 2007
2. Dec 6, 2007

### Shooting Star

There's a minus sign before the 2nd integral.

3. Dec 6, 2007

### danni7070

Ok thanks for that correction. But what about the 2nd integral? How is it done ? The denominator and numerator are switched if you look at my answer and the TI answer.

4. Dec 6, 2007

### Shooting Star

The minus sign makes your answer correct because -log x = log 1/x.

How to get the integral -- well, give me some time.

5. Dec 6, 2007

### Shooting Star

2dx/x(1+x^2) = 2xdx/x^2(1+x^2).

Put v=x^2 => dv=2xdx. The integral becomes dv/v(1+v). That's simple, I hope.

6. Dec 6, 2007

### danni7070

okey. I tried this and although it looks very logical to me this is not correct??

I got int dv/2v = ln(x^2)/2

any ideas?

7. Dec 6, 2007

### Shooting Star

Incorrect. Try the substitution I had shown in my last post.

dv/v(1+v) = [(1+v)-v]dv/v(1+v). Can you complete it?

8. Dec 6, 2007

### danni7070

doh. I see it now. But I still get $$\ln (\frac{x^2}{x^2+1})$$

9. Dec 6, 2007

### Shooting Star

There's a minus sign before this one.

10. Dec 6, 2007

### danni7070

What does that change? I thought that it would be the same result but a negative one.

11. Dec 6, 2007

### danni7070

OK! I get it! It changes everything!

Thank you very much.