Calculus Problem: Simplifying xln(2x+1)-x+1/2ln(2x+1)+C

[Stevecgz]

In the solutions manual to my calculus text I don't understand how they are simplifying the following:

$$x\ln(2x+1)-x+\frac 1 2 \ln(2x+1)+C$$

to...

$$\frac 1 2 (2x+1)\ln(2x+1) -x +C$$

If someone could explain to me how they are simplifying this it would be appreciated.

Steve

 

[Pyrrhus]

First let's group

$$x\ln(2x+1)+\frac{1}{2} \ln(2x+1)-x+C$$

Factor the $\ln(2x+1)$

$$\ln(2x+1)(x +\frac{1}{2})-x+C$$

Multiply by $\frac{2}{2}$

$$\ln(2x+1) \frac{2}{2} (x +\frac{1}{2})-x+C$$

$$\ln(2x+1) \frac{1}{2} (2x +\frac{2}{2})-x+C$$

$$\ln(2x+1) \frac{1}{2} (2x + 1)-x+C$$

 

[Stevecgz]

Thank you much Cyclovenom, I understand now.

Steve