Question #9adb8

1 Answer
Feb 15, 2018

#int(2x+6-6)/(x^2+6x+13)dx=ln(x^2+6x+13)-3tan^-1((x+3)/2)#

Explanation:

The problem
#int(2x)/(x^2+6x+13)dx# can be solved By the method as shown below.
Let #u=x^2+6x+13#, #du=(2x+6)dx#
Further, in the numerator, we have only #2x=2x+6-6#
the problem can be restated as
Integrate the expression
#int(2x+6-6)/(x^2+6x+13)dx#
By applying sum rule,
#int(2x+6-6)/(x^2+6x+13)dx=int(2x+6)/(x^2+6x+13)dx-int(6)/(x^2+6x+13)dx#
Let #I_1=int(2x+6)/(x^2+6x+13)dx#
and
Let #I_2 =int(6)/(x^2+6x+13)dx#

Now,
#int(2x+6-6)/(x^2+6x+13)dx=I_1-I_2#

Evaluating #I_1,#

#I_1=int(2x+6)/(x^2+6x+13)dx=int(du)/u=lnu#

Substituting for #u=x^2+6x+13#

#I_1=ln(x^2+6x+13)#

Evaluating #I_2,#

#I_2 =int(6)/(x^2+6x+13)dx#

Completing the squares in the denominator

#x^2+6x+13=(x+3)^2+2^2#
Simpliying
#x^2+6x+13=2^2(((x+3)/2)^2+1)=4(1+((x+3)/2)^2)#
Substituting
#I_2 =int(6)/(4(1+((x+3)/2)^2))dx=3/2int(1)/(1+((x+3)/2)^2)dx#

Let #(x+3)/2=t#, #dx=2dt#
#I_2 =3/2int(1)/(1+t^2)2dt=3int(1)/(1+t^2)dt=3tan^-1t#
#I_2=3tan^-1t#
Substituting for #t=(x+3)/2#
#I_2=3tan^-1((x+3)/2)#

#int(2x+6-6)/(x^2+6x+13)dx=I_1-I_2#

#I_1=ln(x^2+6x+13)#

#I_2=3tan^-1((x+3)/2)#

#int(2x+6-6)/(x^2+6x+13)dx=ln(x^2+6x+13)-3tan^-1((x+3)/2)#