[tex]\int_{0}^{1}\frac{1}{1+x^2}dx[/tex]
The integral above is definite so we must first calculate for indefinite one.
[tex]\int{\frac{1}{1+x^2}dx}[/tex]
Rule: [tex]\int{\frac{1}{a^2+b^2}dx}=\frac{1}{b}\times\arctan(\frac{a}{b})[/tex].
Now we apply this rule and get:
[tex]\int{\frac{1}{1+x^2}}=\frac{1}{1}\times\arctan(\frac{x}{1})[/tex]
Or just simply: [tex]\arctan(x)[/tex]
Now we integrate:
[tex]\arctan(x)\Big\vert_{0}^{1}[/tex]
[tex]\arctan(1)-\arctan(0)[/tex]
[tex]\frac{\pi}{4}-0\implies\boxed{\frac{\pi}{4}}[/tex]
