Find the area of the enclosed region

The region is enclosed by the curves $y=\sqrt{x+2}$, $y=\frac1{x+1}$, and lies between $x=0$ and $x=2$. Help please, any work would be helpful. Thank you!

• Any work (of your own) would be helpful in understanding what you are stuck on. Have you drawn the region? – Matthew Leingang Dec 14 '14 at 17:05
• Dang, I submitted my edit just after Sujaan's.... – teadawg1337 Dec 14 '14 at 17:06

Integrate the difference between 0 and 2:

f1 = Sqrt[x + 2];
f2 = 1/(x + 1);
Plot[{f1, f2}, {x, 0, 2}, Filling -> {1 -> {{2}, {None, LightGray}}},
PlotTheme -> "Detailed", PlotLegends -> {f1, f2}]


Integrate[f1 - f2, {x, 0, 2}]


N[%]


Hint. Since, for $x$ such that $0\leq x\leq2$, we have $\displaystyle \sqrt{x+2} > \frac1{x+1}$, then the desired area $A$ is such that $$A =\int_0^2 \sqrt{x+2} \:dx-\int_0^2 \frac1{x+1} \:dx$$ The second integral on the right hand side is easy to find out, for the first integral just write $\displaystyle \sqrt{x+2}=(x+2)^{1/2}$.