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

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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)#