# What is the net area between f(x) = x^2-ln(x^2+1)  and the x-axis over x in [1, 2 ]?

Oct 24, 2016

$= \frac{1}{3} - \ln \left({5}^{2} / 2\right) + 2 \arctan 2 - \frac{\pi}{2}$

#### Explanation:

To find the area between $f \left(x\right)$ and x-axis is to find the definite integral of this function between $\left[1 , 2\right]$

${\int}_{1}^{2} f \left(x\right)$

$= {\int}_{1}^{2} {x}^{2} - \ln \left({x}^{2} + 1\right) \mathrm{dx}$
$= \textcolor{b r o w n}{{\int}_{1}^{2} {x}^{2} \mathrm{dx}} - \textcolor{b l u e}{{\int}_{1}^{2} \ln \left({x}^{2} + 1\right) \mathrm{dx}}$

Let us compute the first integral color(brown)(int_1^2x^2dx
As we know integral of the polynomial $\int {x}^{n} = {x}^{n + 1} / \left(n + 1\right)$
So $\textcolor{b r o w n}{\int {x}^{2} \mathrm{dx} = {x}^{3} / 3}$

color(brown)(int_1^2x^2dx=2^3/3-1^3/3=8/3-1/3=7/3

Now, Let us compute the second integral $\textcolor{b l u e}{{\int}_{1}^{2} \ln \left({x}^{2} + 1\right) \mathrm{dx}}$ using integration by parts .

Let
$\textcolor{red}{u \left(x\right) = \ln {x}^{2} + 1} \Rightarrow \mathrm{dc} o l \mathmr{and} \left(| red\right) \left(u \left(x\right) = \frac{2 x}{{x}^{2} + 1} \mathrm{dx}\right)$
then $\textcolor{p u r p \le}{\mathrm{dv} = \mathrm{dx}} \Rightarrow \textcolor{p u r p \le}{v \left(x\right) = x}$

$\textcolor{b l u e}{\int \ln \left({x}^{2} + 1\right) \mathrm{dx}}$
$= \textcolor{red}{u \left(x\right)} \cdot \textcolor{p u r p \le}{v \left(x\right)} - \int \textcolor{p u r p \le}{v \left(x\right)} \cdot \textcolor{red}{\mathrm{du} \left(x\right)}$

$\int \textcolor{p u r p \le}{v \left(x\right)} \cdot \textcolor{red}{\mathrm{du} \left(x\right)}$

$= \int x \cdot \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

$= \int \frac{2 {x}^{2}}{{x}^{2} + 1} \mathrm{dx}$

$= 2 \int \frac{{x}^{2}}{{x}^{2} + 1} \mathrm{dx}$
$= 2 \int 1 - \frac{1}{{x}^{2} + 1} \mathrm{dx}$
$= 2 \int \mathrm{dx} - 2 \int \frac{1}{{x}^{2} + 1} \mathrm{dx}$

$= \textcolor{\mathmr{and} a n \ge}{2 x - 2 \arctan x}$

So,
$\textcolor{b l u e}{\int \ln \left({x}^{2} + 1\right) \mathrm{dx}}$ $= \textcolor{red}{u \left(x\right)} \cdot \textcolor{p u r p \le}{v \left(x\right)} - \int \textcolor{p u r p \le}{v \left(x\right)} \cdot \textcolor{red}{\mathrm{du} \left(x\right)}$

$= \textcolor{p u r p \le}{x} \textcolor{red}{\ln \left({x}^{2} + 1\right)} - \left(\textcolor{\mathmr{and} a n \ge}{2 x - 2 \arctan x}\right)$

$= x \ln \left({x}^{2} + 1\right) - 2 x + 2 \arctan x$

$\textcolor{b l u e}{{\int}_{1}^{2} \ln \left({x}^{2} + 1\right) \mathrm{dx}}$
$= 2 \ln \left({2}^{2} + 1\right) - 2 \cdot 2 + 2 \arctan 2 - \left(1 \cdot \ln \left({1}^{2} + 1\right) - 2 \cdot 1 + 2 \arctan 1\right)$

$= 2 \ln 5 - 4 + 2 \arctan 2 - \left(\ln 2 - 2 + 2 \arctan 1\right)$
$= 2 \ln 5 - 4 + 2 \arctan 2 - \ln 2 + 2 - 2 \cdot \frac{\pi}{4}$
$= 2 \ln 5 - \ln 2 - 4 + 2 + 2 \arctan 2 - \frac{\pi}{2}$
$= \ln \left({5}^{2}\right) - \ln 2 - 2 + 2 \arctan 2 - \frac{\pi}{2}$
color(blue)(=ln(5^2/2)-2+2arctan2-pi/2

${\int}_{1}^{2} f \left(x\right)$

$= {\int}_{1}^{2} {x}^{2} - \ln \left({x}^{2} + 1\right) \mathrm{dx}$
$= \textcolor{b r o w n}{{\int}_{1}^{2} {x}^{2} \mathrm{dx}} - \textcolor{b l u e}{{\int}_{1}^{2} \ln \left({x}^{2} + 1\right) \mathrm{dx}}$
=color(brown)(7/3)-color(blue)(ln(5^2/2)-2+2arctan2-pi/2
=color(brown)(7/3)-color(blue)(ln(5^2/2)-2+2arctan2-pi/2
$= \frac{7}{3} - 2 - \ln \left({5}^{2} / 2\right) + 2 \arctan 2 - \frac{\pi}{2}$
$= \frac{7}{3} - \frac{6}{3} - \ln \left({5}^{2} / 2\right) + 2 \arctan 2 - \frac{\pi}{2}$

$= \frac{1}{3} - \ln \left({5}^{2} / 2\right) + 2 \arctan 2 - \frac{\pi}{2}$