Question

Evaluate the definite integrals: $\int^1_0 {\frac{dx}{1+x^2}$

Answer 1

Answer:

$\frac{\pi}{4}$

Step-by-step explanation:

Concept:

$1.\int{\frac{1}{1+x^2}}\:dx=tan^{-1}x+c\\\\2.tan\frac{\pi}{4}=1\\\\3.tan0=0$

Now,

$\int\limits^1_0{\frac{1}{1+x^2}}\:dx\\\\=[tan^{-1}x]^1_0\\\\=tan^{-1}(1)-tan^{-1}(0)\\\\=\frac{\pi}{4}-0\\\\=\frac{\pi}{4}$