Calculating area using integrals: y = 1/(1+x^2), y = 0, x = 0, x = 1

[Johulus]

The task which I am currently trying to solve requires to determine the area encircled by following curves:

\(\displaystyle y=\dfrac{1}{1+x^2} \qquad x=0 \qquad x=1 \qquad y=0 \)

I should be able to calculate the required area using integrals as follows:

\(\displaystyle A=\int_a^b \mathrm{f(x)} \, \mathrm{d}x=\int_0^1 \mathrm{\dfrac{1}{1+x^2}} \, \mathrm{d}x \)

But the problem is I don't know how to get antiderivative from the given expression. I tried:

\(\displaystyle \int \mathrm{\dfrac{1}{1+x^2}} \, \mathrm{d}x=\int (\mathrm{\dfrac{1+x^2}{1+x^2}-\dfrac{x^2}{1+x^2}}) \, \mathrm{d}x=\int (\mathrm{1-\dfrac{x^2}{1+x^2}}) \, \mathrm{d}x=\int \mathrm{d}x - \int \mathrm{\dfrac{x^2}{1+x^2}} \, \mathrm{d}x=x- \dfrac{1}{2} \int \mathrm{\dfrac{2x^2}{1+x^2}} \, \mathrm{d}x= x- \dfrac{1}{2}x \ln{(1+x^2)} + \dfrac{1}{2} \int \mathrm{\ln{(1+x^2)}} \, \mathrm{d}x \)

And at this point I don't know what to do with \(\displaystyle \int \mathrm{\ln{(1+x^2)}} \, \mathrm{d}x \) .

Could someone give me some hint?

 

[Otis]

The antiderivative of 1/(1+x^2) is a standard one. If we can't remember it, we can use a trigonometric substitution to calculate it.

Let x = tan(θ), so θ = arctan(x)

Then 1/(1+x^2) dx becomes 1/[1 + tan^2(θ)] dθ

Also, dx/dθ = sec^2(θ)

And, this identity is useful: 1 + tan^2(θ) = sec^2(θ)

Try those hints. :cool: