# Differentiation of Ln(x+1/x)

1. Nov 2, 2008

### jorgen

Hi all,

I have to differentiate

$$Ln(x+\frac{1}{x})$$

where I first differentiate Ln and than multiply by the differentiation of the inner function

$$1/(x+\frac{1}{x})*(1-\frac{1}{x^2}$$

which I simplify to

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

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

the problem is I cannot rewrite it to this

$$\frac{2*x}{x^2+1}-\frac{1}{x}$$

how to rewrite it - any help or advise appreciated. Thanks in advance

Best
Jorgen

Last edited: Nov 2, 2008
2. Nov 2, 2008

### borgwal

Start at the final expression, and write it as a fraction with one denominator: x(x^2+1)

3. Nov 2, 2008

### jorgen

thanks,

so I put into one fraction

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

but I don't know how to start rearranging this.... Any new hints

Best

J

4. Nov 2, 2008

### gabbagabbahey

$x^2-1=(x^2+1)-2$ ;0)

5. Nov 2, 2008

### jorgen

so I rewrite the fraction using this hint

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

I split the fraction into

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

but I can still not see how to rearrange it

Best
J

6. Nov 2, 2008

### gabbagabbahey

Use partial fractions to decompose your second term; in other words, find the constants $A$,$B$, and $C$ that satisfy:

$$\frac{2}{x*(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$$

7. Nov 3, 2008

### Tedjn

In what form do you want it? Whether it is condensed into one fraction or written as a difference doesn't matter if both expressions are equal.