<s>[INST] Prove that $2<\int^2_1ln(x^2+1)+x<3$ To show that $2<\int^2_1ln(x^2+1)+x$ , I took the derivative of the integrand and got that it's positive always in $[1,2]$, so its strictly increasing and the minimum in $[1,2]$ is at $x=1$, so $ln(2)+1<ln(x^2+1)+x$, and $ln(2)<ln(e)=1$ so: $ln(2)+1<2< ln(x^2+1)+x$ in $(1,2]$. 
EDIT: I just realized that the lower bound doesn't work too. 
About the upper bound I tried to substitute $2$ but it becomes $ln(5)+2>3$, so it doesn't really help. 
I would appreciate any help.  Thanks in advance.
 [/INST] I began to solve this before Bacha's solution appeared, so I did not delete.
Hope I'm correct.
$$\int\ln(x^2+1)dx=(*)\\let\:f=\ln(x^2+1),g=x\\then\:df=\frac{2x}{x^2+1}dx,dg=dx\\\text{using integrating by parts where }\int fdg=dg-\int gdf\text{ we get}\\(*)=x\ln(x^2+1)-\int\frac{2x^2}{x^2+1}dx=x\ln(x^2+1)-2\int \left(1-\frac{1}{x^2+1}\right)dx=\\=x\ln(x^2+1)-2x+2\arctan(x)+C$$
And now calculate
$$\int\limits_1^2(\ln(x^2+1)+x)dx=\\=2\ln(2^2+1)-2*2+2\arctan(2)-\ln(1^2+1)+2*1-2\arctan(1)+\frac{1}{2}2^2-\frac{1}{2}1^2=\\=\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)$$
If we use calculator then $\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)\approx2.6692$. But if you want a more "honest" solution then we can try to estimate this from left and from right but it would hard... Nevertheless let's see how estimating would look.
From left:
$$\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)>\ln(e^2)-\frac{1}{2}-\frac{\pi}{2}+2\arctan(\sqrt{3})=2-\frac{1}{2}-\frac{\pi}{2}+\frac{2\pi}{3}=\\=1.5+\frac{\pi}{6}> 2$$
Ok.
From right:
$$\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)<\\<\left(\ln(e^2)+(\frac{25}{2}-e^2)*\frac{1}{e^2}\right)^{[2]}-\frac{1}{2}-\frac{\pi}{2}+2*\left(\arctan(\sqrt 3)+(2-\sqrt 3)*\frac{1}{\sqrt 3^2+1}\right)^{[1]}<\\<\left(2+5.5*\frac{1}{7}\right)-\frac{1}{2}-\frac{\pi}{2}+\left(\frac{2\pi}{3}+(2-\sqrt 3)*\frac{1}{2}\right)<\\<2.8-\frac{1}{2}-\frac{\pi}{2}+\frac{2\pi}{3}+(2-\sqrt 3)*\frac{1}{2}=2.3+\frac{\pi}{6}+(2-\sqrt 3)*\frac{1}{2}<2.3+\frac{\pi}{6}+0.3*\frac{1}{2}=\\=2.45+\frac{\pi}{6}<3$$
[1] $\forall x\geq 0:\:\arctan(x) + (x+\delta x)*\arctan'(x)\geq\arctan(x+\delta x)$ because $\forall y>x\geq 0:\: \arctan'(x)>\arctan'(y)$.
[2] Same as [1].
Ok, but too much work.
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