Can someone please integrate this (hard) (not sure if even possible)

[trot]

Please intergrate: (x+3)/(x^2+4x) is it even possible to integrate this?

The question in the book was (x+2)/(x^2+4x) and it can nicely be integrated to 0.5ln(x^2+4x), but by changing the 2 to a 3 f'(x) don't go into each other. if you know what i mean.

 

[DonAntonio]

Please intergrate: (x+3)/(x^2+4x) is it even possible to integrate this?

The question in the book was (x+2)/(x^2+4x) and it can nicely be integrated to 0.5ln(x^2+4x), but by changing the 2 to a 3 f'(x) don't go into each other. if you know what i mean.

First:

$$\frac{1}{x^2+4x}=\frac{1}{4}\left[\frac{1}{x}-\frac{1}{x+4}\right]$$,

so that

$$\int\frac{x+3}{x^2+4x}=\frac{1}{2}\int\frac{2x+4}{x^2+4x}dx+\int\frac{1}{x^2+4x}dx=\frac{1}{2}\log(x^2+4x)+\frac{1}{4}\log\left(\frac{x}{x+4}\right)+C$$

 

[Ray Vickson]

First:

$$\frac{1}{x^2+4x}=\frac{1}{4}\left[\frac{1}{x}-\frac{1}{x+4}\right]$$,

so that

$$\int\frac{x+3}{x^2+4x}=\frac{1}{2}\int\frac{2x+4}{x^2+4x}dx+\int\frac{1}{x^2+4x}dx=\frac{1}{2}\log(x^2+4x)+\frac{1}{4}\log\left(\frac{x}{x+4}\right)+C$$

You are not supposed to do a student's homework---just give hints.

RGV

 

[micromass]

You are not supposed to do a student's homework---just give hints.

RGV

Right. Please just give hints next time and don't solve the entire problem (even if it's not posted in the homework section: if it looks like homework, then don't give full solutions).