How do you evaluate #int# #sqrt(x^2 + 14x)# dx?

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

#int# #sqrt(x^2 + 14x)# dx?

1 Answer
Narad T.
Jun 17, 2017

The answer is #=1/2(x+7)sqrt(x^2+14x)-49/2ln(|1/7(sqrt(x^2+14x)+(x+7))|)+C#

Explanation:

We need

#cosh^2theta-sinh^2theta=1#

#cosh 2theta=2sinh^2theta+1#

#sinh^2theta=1/2(cosh 2theta-1)#

#sinh2theta=2sinhthetacoshtheta#

#arc cosh x=ln(sqrt(x^2-1)+x)#

We perform this integral by substitution but first we do some simplification

#x^2+14x=x^2+14x+49-49=(x+7)^2-49#

The substitution is

#x+7=7coshtheta#

#dx=7sinhtheta d theta#

#sqrt((x+7)^2-49)=sqrt(49cosh^2theta-49)#

#=7sqrt(cosh^2theta-1)#

#=7sinh theta#

#Sinh^2 theta=cosh^2theta-1=((x+7)/7)^2-1#

#=(x^2+14x+49-49)/49#

#=(x^2+14x)/49#

#sinh theta=1/7sqrt(x^2+14x)#

Therefore,

#intsqrt(x^2+14x)dx=int7sinhtheta*7sinh theta d theta#

#=49int sinh^2 theta d theta#

#=49/2int(cosh 2theta-1) d theta#

#=49/2((sinh 2theta)/2-theta)#

#=49/2(sinhthetacoshtheta-arc cosh((x+7)/7))#

#=(49/2*1/7*sqrt(x^2+14x)*(x+7)/7)-49/2(ln(sqrt((x+7)^2/49-1))+(x+7)/7)+C#

#=1/2(x+7)sqrt(x^2+14x)-49/2ln(|1/7(sqrt(x^2+14x)+(x+7))|)+C#

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