#int1/sqrt(3x-12sqrtx+53)dx#
#=1/sqrt3int1/sqrt(x-4sqrtx+53/3)dx#
This form of integral doesn't seem to be know: let substitute:
Let
#u=sqrtx#
#u²=x#
#dx=2udu#
So:
#1/sqrt3int1/sqrt(x-4sqrtx+53/3)dx=1/sqrt3int(2u)/sqrt(u²-4u+53/3)du#
#=2/sqrt3intu/sqrt(u²-4u+53/3)du#
Now we need to complete the square :
#=2/sqrt3intu/sqrt((u-2)²+41/3)du#
Now, let : #u-2=sqrt(41/3)cot(theta)#
#du=-sqrt(41/3)csc(theta)²d theta#
So :
#2/sqrt3int(u*du)/sqrt(u²-4u+53/3)#
#=2/sqrt3int((sqrt(41/3)cot(theta)+2)(-sqrt(41/3)csc(theta)²))/sqrt(41/3cot(theta)²+41/3)d theta#
#=-2/sqrt3int((sqrt(41/3)cot(theta)+2)(cancel(sqrt(41/3))csc(theta)²))/(cancel(sqrt(41/3))sqrt(cot(theta)²+1))d theta#
Because #cot(theta)²+1=csc(theta)²#,
#-2/sqrt3int((sqrt(41/3)cot(theta)+2)csc(theta)²)/(sqrt(cot(theta)²+1))d theta=-2/sqrt3int((sqrt(41/3)cot(theta)+2)csc(theta)^(cancel(2)))/cancel(csc(theta))d theta#
#=-2/sqrt3intsqrt(41/3)cot(theta)csc(theta)d theta-2/sqrt3intcsc(theta)d theta#
#=-2/3sqrt41intcos(theta)/(sin(theta)²)d theta-2/sqrt3intcsc(theta)d theta#
Now we've got two easier integrals : I won't come back on #intcsc(theta)=-ln(csc(theta)+cot(theta))#, you can see there the demonstration.
For the second one, remarkably, #(-cos(x))^'=sin(x)#
So: #-intcos(theta)/(sin(theta)²)d theta=-int(f^')/(f²)=1/f=1/sin(theta)#
So: #-2/3sqrt41intcos(theta)/(sin(theta)²)d theta-2/sqrt3intcsc(theta)d theta#
#=(2sqrt41)/(3sin(theta))+2/sqrt3ln(csc(theta)+cot(theta))#
But #theta=cot^(-1)(sqrt(3/41)(u-2))#
So:
#(2sqrt41)/(3sin(theta))+2/sqrt3ln(csc(theta)+cot(theta))#
#=(2sqrt41)/(3sin(cot^(-1)(sqrt(3/41)(u-2))))+2/sqrt3ln(|csc(cot^(-1)(sqrt(3/41)(u-2)))+sqrt(3/41)(u-2)|)#
Finally, #u=sqrtx#
So:
#int1/sqrt(3x-12sqrtx+53)dx#
#=(2sqrt41)/(3sin(cot^(-1)(sqrt(3/41)(sqrtx-2))))+2/sqrt3ln(|csc(cot^(-1)(sqrt(3/41)(sqrtx-2)))+sqrt(3/41)(sqrtx-2)|)+C# #C in RR#
\0/ Here's our answer !
