Tutorial Exercise

Evaluate the integral.

${\displaystyle \underset{0}{\overset{1}{}}}\frac{3}{t^2+1}dt$

Part $1$ of $4$

To find an antiderivative for $f(t)=\frac{3}{1+t^2},$ think about the inverse trigonometric functions and remember that $\frac{1}{1+t^2}=\frac{d}{dt}[].$

Therefore, an antiderivative for $f(t)=\frac{3}{1+t^2}$ is $F(t)=$

Part $2$ of $4$

We now have ${\displaystyle \underset{0}{\overset{1}{}}}\frac{3}{1+t^2}dt=[3arctan(t){]}_0^1=3arctan(1)-3arctan(0).$

We know that $arctan(1)$ equals the angle between $-\frac{}{2},\frac{}{2}$ whose tangent is $1.$

This angle is

radians.