#intsqrt(x^2+25) dx# ?

Question

766 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-26T14:12:13

The answer is #=25/2arcsinh(x/5)+1/2xsqrt(25+x^2)+C#

Perform a hyperbolic substitution

Let #x=5sinhu#, #=>#, #dx=5coshu du#

#sqrt(x^2+25)=sqrt(25sinh^2u+25)=5coshu#

#cosh(2u)=2cosh^2u-1#

#cosh^2u=(cosh(2u)+1)/2#

#sinh(2u)=2sinhucoshu=2/5xsqrt(1+(x/5)^2)#

Therefore, the integral is

#intsqrt(x^2+25)dx=int5coshu*5coshudu#

#=25intcosh^2udu#

#=25/2int(cosh(2u)+1)du#

#=25/2(1/2sinh(2u)+u)#

#=25/4sinh(2u)+25/2u#

#=25/2arcsinh (x/5)+25/4*2/5xsqrt(1+x^2/25)+C#

#=25/2arcsinh(x/5)+1/2xsqrt(25+x^2)+C#