Use trigonometric substitution.
Draw a triangle with an angle #theta#.
Label the opposite side as #x# and adjacent side as #5#.
Now examine the triangle and notice that
#tantheta=x/5#
#rArrx=5tantheta#
#rArrdx=5sec^2theta# #d##theta#
By the Pythagorean Theorem, the hypotenuse of the triangle is:
#sqrt(25+x^2)#
So we can write
#sectheta=sqrt(25+x^2)/5#
#rArrsqrt(25+x^2)=5sectheta#
Let's now rewrite the integral in terms of #theta#
#intsqrt(25+x^2)dxrArrint(5sectheta)(5sec^2theta# #d##theta)#
#rArr25int# #sec^3theta# #d##theta#
Integrating #sec^3theta# can be tricky. There is a shortcut formula, but I'll do it the long way for teaching purposes.
Separate #(sec^3theta# #d##theta)# into #(sectheta*sec^2theta# #d##theta)# and use integration by parts.
Let:
#u=sectheta#
#du=secthetatantheta# #d##theta#
#dv=sec^2theta# #d##theta#
#v=tantheta#
Then:
#rArrint# #sec^3theta# #d##theta=secthetatantheta-int# #tan^2thetasectheta# #d##theta#
#rArrint# #sec^3theta# #d##theta=secthetatantheta-int# #(sec^2theta-1)sectheta# #d##theta#
#rArrint# #sec^3theta# #d##theta=secthetatantheta-int# #(sec^3theta-sectheta)##d##theta#
#rArr2int# #sec^3theta# #d##theta=secthetatantheta+int# #sectheta# #d##theta#
#rArr2int# #sec^3theta# #d##theta=secthetatantheta+lnabs(sectheta+tantheta)#
#rArrint# #sec^3theta# #d##theta=1/2(secthetatantheta+lnabs(sectheta+tantheta))+C#
And let's not forget that our integral was multiplied by 25!
#rArr25int# #sec^3theta# #d##theta=25/2(secthetatantheta+lnabs(sectheta+tantheta))+C#
Now put everything back in terms of #x#
#rArr25/2(sqrt(25+x^2)/5x/5+lnabs(sqrt(25+x^2)/5+x/5))+C#
#rArr1/2(xsqrt(25+x^2)+25lnabs((sqrt(25+x^2)+x)/5))+C#
#rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2))-25ln5)+C#
Then we can absorb the constant #-(25ln5)/2# into #C# to get our final answer:
#rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2)))+C#
