intdx/sqrt(3x-12sqrtx-21)∫dx√3x−12√x−21
I've added the differential dxdx to your integrand. While it may seem like a fanciful calculus accessory, it's actually very important, especially when doing trig sub problems.
First, let's get this into a more "normal" form for trig subs. Let x=t^2x=t2. Note this implies that dx=2tdtdx=2tdt and that sqrtx=t√x=t.
=int(2tdt)/sqrt(3t^2-12t-21)=∫2tdt√3t2−12t−21
Now complete the square in the denominator:
=2intt/sqrt(3(t^2-4t+4)-21-12)dt=2∫t√3(t2−4t+4)−21−12dt
=2intt/sqrt(3(t-2)^2-33)dt=2∫t√3(t−2)2−33dt
=2/sqrt3intt/sqrt((t-2)^2-11)dt=2√3∫t√(t−2)2−11dt
This is optional, but now I'd let s=t-2s=t−2. This implies that ds=dtds=dt and that t=s+2t=s+2.
=2/sqrt3int(s+2)/sqrt(s^2-11)ds=2√3∫s+2√s2−11ds
Now, to clear the denominator, I'd let s=sqrt11secthetas=√11secθ.
My reason for this substitution is that s^2-11=11sec^2theta-11=11(sec^2theta-1)=11tan^2thetas2−11=11sec2θ−11=11(sec2θ−1)=11tan2θ. This will clear out our denominator.
Moreover, note that ds=sqrt11secthetatanthetad thetads=√11secθtanθdθ. Proceeding:
=2/sqrt3int(sqrt11sectheta+2)/sqrt(11tan^2theta)(sqrt11secthetatanthetad theta)=2√3∫√11secθ+2√11tan2θ(√11secθtanθdθ)
=2/sqrt3int(sqrt11sectheta+2)/(sqrt11tantheta)(sqrt11secthetatanthetad theta)=2√3∫√11secθ+2√11tanθ(√11secθtanθdθ)
=2/sqrt3int(sqrt11sectheta+2)secthetad theta=2√3∫(√11secθ+2)secθdθ
=2/sqrt3int(sqrt11sec^2theta+2sectheta)d theta=2√3∫(√11sec2θ+2secθ)dθ
These are two fairly well known integrals. You may need to look the second one up.
=2/sqrt3(sqrt11tantheta+2ln(abs(sectheta+tantheta)))=2√3(√11tanθ+2ln(|secθ+tanθ|))
Recall that s=sqrt11secthetas=√11secθ, so sectheta=s/sqrt11secθ=s√11. Moreover, tantheta=sqrt(sec^2theta-1)=sqrt(s^2/11-1)=sqrt((s^2-11)/11)tanθ=√sec2θ−1=√s211−1=√s2−1111. Then:
=2sqrt(11/3)sqrt((s^2-11)/11)+4/sqrt3ln(abs(s/sqrt11+sqrt((s^2-11)/11)))=2√113√s2−1111+4√3ln(∣∣∣s√11+√s2−1111∣∣∣)
Do some more simplifications. Remember the log rule log(A//B)=log(A)-log(B)log(A/B)=log(A)−log(B).
=2/sqrt3sqrt(s^2-11)+4/sqrt3ln(abs(s+sqrt(s^2-11)))-4/sqrt3lnsqrt11=2√3√s2−11+4√3ln(∣∣s+√s2−11∣∣)−4√3ln√11
Now use s=t-2s=t−2 and t=sqrtxt=√x, so s=sqrtx-2s=√x−2;
=2/sqrt3((sqrtx-2)^2-11)+4/sqrt3ln(abs(sqrtx-2+sqrt((sqrtx-2)^2-11)))-4/sqrt3lnsqrt11=2√3((√x−2)2−11)+4√3ln(∣∣∣√x−2+√(√x−2)2−11∣∣∣)−4√3ln√11
=2/sqrt3(x-4sqrtx-7)+4/sqrt3ln(sqrtx+2+sqrt(x-4sqrtx-7))+C=2√3(x−4√x−7)+4√3ln(√x+2+√x−4√x−7)+C
I've added the constant of integration, into which the 4/sqrt3lnsqrt114√3ln√11 term has been absorbed.
The absolute value bars are also unnecessary because the argument of the natural log function is, by necessity, greater than 22 and always positive.
