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

##### 1 Answer
Apr 30, 2017

The net area will be $- 1.17$ square units.

#### Explanation:

Start by finding the antiderivative of $f \left(x\right)$, call it $g \left(x\right)$.

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

Use integration by parts. Let $u = \ln \left({x}^{2} + 1\right)$ and $\mathrm{dv} = 1 \mathrm{dx}$. Then $\mathrm{du} = \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$ and $v = x$.

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$\int \ln \left({x}^{2} + 1\right) = \ln \left({x}^{2} + 1\right) x - \int x \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

$\int \ln \left({x}^{2} + 1\right) = x \ln \left({x}^{2} + 1\right) - 2 \int {x}^{2} / \left({x}^{2} + 1\right) \mathrm{dx}$

$\int \ln \left({x}^{2} + 1\right) = x \ln \left({x}^{2} + 1\right) - 2 \int 1 \mathrm{dx} + 2 \int \frac{1}{{x}^{2} + 1} \mathrm{dx}$

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

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

Now let's consider the region whose area we must determine.

We now set up a definite integral.

${\int}_{1}^{2} \ln \left({x}^{2} + 1\right)$. This will make:

$A = 2 \left(2\right) - 2 \arctan \left(2\right) - 2 \ln \left(5\right) - \left(2 - 2 \arctan \left(1\right) - \ln \left(2\right)\right)$

$A \approx - 1.17$

Hopefully this helps!