How do you evaluate $int dx/(x^2+4x+13)$ from $[-2, 2]$ ?

Question

12363 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-24T16:27:33

$int_(-2)^2dx/(x^2+4x+13)=1/3arctan(4/3)$

First without the bounds:

$I=intdx/(x^2+4x+13)$

Complete the square in the denominator.

$I=intdx/((x+2)^2+9)$

Now we can use trigonometric substitution. Let $x+2=3tantheta$ . This also implies that $dx=3sec^2thetad theta$ .

$I=int(3sec^2thetad theta)/(9tan^2theta+9)=1/3int(sec^2thetad theta)/(tan^2theta+1)=1/3int(sec^2thetad theta)/sec^2theta$

$I=1/3intd theta=1/3theta$

From $x+2=3tantheta$ we see that $theta=arctan((x+2)/3)$ .

$I=1/3arctan((x+2)/3)$

So now applying the bounds:

$I_B=int_(-2)^2dx/(x^2+4x+13)=[1/3arctan((x+2)/3)]_(-2)^2$

So

$I_B=(1/3arctan((2+2)/3))-(1/3arctan((-2+2)/3))$

$I_B=1/3arctan(4/3)$

How do you evaluate int dx/(x^2+4x+13)int dx/(x^2+4x+13) from [-2, 2][-2, 2]?