Area between $5 \, x \, \ln(x)$ and $(x+2) \, \ln(x)$ Can someone help me with this?
I have already found their intersections
$$x = \frac{1}{2} \hspace{4mm} \text{and} \hspace{4mm} x = 1.$$
I did this:
$$\int_{\frac{1}{2}}^{1} (5x\ln(x) - (x+2)\ln(x)) \, \mathrm d x$$
I conclude in:
$$ 2 \int_{\frac{1}{2}}^{1} ((x^2 - x) \, \ln(x) - (x^2 -x)) \, \mathrm d x $$
and I get $$ \frac{\ln({\frac{1}{2}}) - 1}{2}$$
but it's not the same that wolfram alpha shows here.
 A: Well, the two intersections can be found by solving:
$$5x\ln\left(x\right)=\left(x+2\right)\ln\left(x\right)\space\Longleftrightarrow\space x=\frac{1}{2}\space\space\space\bigvee\space\space\space x=1\tag1$$
So, now in order to find the area between the two curves:


*

*$$\mathcal{A}_1=\int_\frac{1}{2}^15x\ln\left(x\right)\space\text{d}x=\frac{5\left(\ln\left(4\right)-3\right)}{16}\tag2$$

*$$\mathcal{A}_2=\int_\frac{1}{2}^1\left(x+2\right)\ln\left(x\right)\space\text{d}x=\frac{19}{16}-\frac{9\ln2}{8}\tag3$$


So, we get:
$$\mathcal{A}=\mathcal{A}_2-\mathcal{A}_1=\frac{19}{16}-\frac{9\ln2}{8}-\frac{5\left(\ln\left(4\right)-3\right)}{16}=\frac{\ln\left(4\right)-1}{4}\tag4$$
A: Well, it is seen from the graph that $(x+2)\ln x \geq 5x\ln x$ over $[0.5,1]$. Thus we have to find, $$\int_{0.5}^{1} (x+2)\ln x - 5\ln x \mathrm{d}x$$ $$= \int_{0.5}^{1} 2\ln x - 4x\ln x \mathrm{d}x$$ $$=\int_{0.5}^{1} -2(2x-1)\ln x \mathrm{d}x$$ $$= [-2(x^2-x)\ln x]|_{0.5}^{1} + 2\int_{0.5}^{1} (x-1) \mathrm{d}x$$ $$=\frac{1}{2}[\ln 2 -\frac{1}{2}] = \frac{1}{4}[\ln 4-1]$$
Hope it helps.
