Question

Who helps me with these integral calculus exercises?

Answer 1

$\int_{0}^{1}\frac{1}{1+x^2}dx$

The integral above is definite so we must first calculate for indefinite one.

$\int{\frac{1}{1+x^2}dx}$

Rule: $\int{\frac{1}{a^2+b^2}dx}=\frac{1}{b}\times\arctan(\frac{a}{b})$.

Now we apply this rule and get:

$\int{\frac{1}{1+x^2}}=\frac{1}{1}\times\arctan(\frac{x}{1})$

Or just simply: $\arctan(x)$

Now we integrate:

$\arctan(x)\Big\vert_{0}^{1}$

$\arctan(1)-\arctan(0)$

$\frac{\pi}{4}-0\implies\boxed{\frac{\pi}{4}}$