Respuesta :

r3t40
[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]

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