Minus sign before the 2nd integral

[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:

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

 

[Shooting Star]

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

 

[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.

 

[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.

 

[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.

 

[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?

 

[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?

 

[danni7070]

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

 

[Shooting Star]

There's a minus sign before this one.

 

[danni7070]

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

 

[danni7070]

OK! I get it! It changes everything!

Thank you very much.