MHB What is the Mystery Behind the Extra 1/2 in This Integration?

[karush]

$$\int\frac{dy}{y\left(1+\ln\left(y^2\right)\right)}=$$

using $$\int\frac{1}{u}du = \ln\left({u}\right)+C$$

so then $$\ln\left({y\left(1+ln\left(y^2\right)\right)}\right) + C$$

my TI-Inspire cx cas returned as the answer
$$\frac{\ln\left({\ln\left({y^{2}}\right)}+1\right)}{2} +C$$

but don't know where the $\frac{1}{2}$ came from

 

[MarkFL]

I would begin by rewriting the integral as:

$$\frac{1}{2}\int\frac{1}{\ln\left(y^2\right)+1}\cdot\frac{2}{y}\,dy$$

Now, let:

$$u=\ln\left(y^2\right)+1\,\therefore\,du=\frac{2y}{y^2}\,dy=\frac{2}{y}\,dy$$

Thus, we have:

$$\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|+C$$

The rest is just back-substitution...do you now see where the $$\frac{1}{2}$$ is necessary?

 

[karush]

yes, great help. I should of rewritten the $\int$