Computing Definite Integrals — Question 6

Question 6 - Solution

First observe that the function $$\frac{1}{1+x^2}$$ is an even function, since $$\frac{1}{1+(-x)^2}=\frac{1}{1+x^2}.$$

Therefore, the integral over the symmetric interval $[-2,2]$ can be written as $$\int_{-2}^{2} \frac{1}{1+x^2}\,dx = 2\int_{0}^{2} \frac{1}{1+x^2}\,dx.$$

Now evaluate the integral. Recall that $$\int \frac{1}{1+x^2}\,dx=\arctan x.$$

Thus, $$2\int_{0}^{2} \frac{1}{1+x^2}\,dx = 2\left[\arctan x\right]_{0}^{2}.$$

Evaluate the bounds: $$\arctan(2)-\arctan(0)=\arctan(2).$$

Therefore, $$\int_{-2}^{2} \frac{1}{1+x^2}\,dx = \boxed{2\arctan(2)}.$$