<s>[INST] Equality of integrals: $ \int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $ In Street-Fighting Mathematics (page 16), Prof. Sanjoy Mahajan states that $$ \displaystyle\int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \displaystyle\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $$ 
I realize this is equivalent to saying that 
$$\displaystyle\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x = \displaystyle\int_{1}^{\infty} \frac  {1}{1+x^2} \, \mathrm{d}x, $$ 
but why  is this true?
 [/INST] I assume you mean
$$\int_0^{\infty} \dfrac{dx}{1+x^2} = 2 \int_0^1 \dfrac{dx}{1+x^2}$$
This is true because
$$\underbrace{\int_1^{\infty} \dfrac{dx}{1+x^2} = \int_1^0 \dfrac{-du/u^2}{1+1/u^2}}_{x = 1/u} = \int_0^1 \dfrac{du}{u^2+1}$$
Hence,
$$\int_0^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} \dfrac{dx}{1+x^2} + \int_1^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} \dfrac{dx}{1+x^2} + \int_0^1 \dfrac{dx}{1+x^2} = 2 \int_0^1 \dfrac{dx}{1+x^2}$$
</s>