We can begin by taking #x^2-14x+45# an re writing it in in completed square/ vertex form giving us:
#(x-7)^2-4# Putting this into the integral will give:
#int(4x)/sqrt((x-7)^2-4)dx#
We can factor the #4# out to the front then play with the numerator a little and we get:
#=4int(x-7+7)/sqrt((x-7)^2-4)dx#
#=4int(x-7)/sqrt((x-7)^2-4)+7/sqrt((x-7)^2-4)dx#
Now lets consider the substitution (we'll have to use a hyperbolic function, not a trig function): #2cosh(u)=x-7#
#2sinh(u)du=dx#
Now putting this substitution in we get:
#4int((2cosh(u))/sqrt(4cosh^2(u)-4)+7/sqrt(4cosh^2(u)-4))sinh(u)du#
Simplify this a little by factoring the #4# from the square root:
#=4int(sinh(u)cosh(u))/sqrt(cosh^2(u)-1)+(7sinh(u))/(2sqrt(cosh^2(u)-1))du#
Now at this point we can use the identity: #cosh^2(u)-1=sinh^2(u)# and then do a bit of cancelling.
That will give us:
#=4int(sinh(u)cosh(u))/sqrt(sinh^2(u))+(7sinh(u))/(2sqrt(sinh^2(u)))du#
#=4int(sinh(u)cosh(u))/sinh(u)+(7sinh(u))/(2sinh(u))du#
#=4intcosh(u)+7/2du#
Now evaluate the integral:
#4(sinh(u)+7/2u)+C=4sinh(u)+14u+C#
Now we can use the identity that we used earlier to convert #sinh# into #cosh#:
#=4(sqrt(cosh^2(u)+1)+7/2u)+C#
Now reverse the substitution and we get:
#4sqrt(((x-7)/2)^2+1)+7cosh^-1((x-7)/2)+C#
At this point you can re write the function in terms of its exponential and logarithmic definitions but I believe this maybe satisfactory>
