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 !