Let's rewrite it like this:
int (x^2)/(sqrt(x^2-4))dx∫x2√x2−4dx
With this, a^2 = 4a2=4 so a = 2a=2. We have the form:
sqrt(x^2 - a^2)√x2−a2
which looks like:
sqrt((asectheta)^2 - a^2) = sqrt(a^2sec^2theta - a^2) = asqrt(sec^2theta-1) = atantheta = 2tantheta√(asecθ)2−a2=√a2sec2θ−a2=a√sec2θ−1=atanθ=2tanθ
So:
Let x = asectheta = 2secthetax=asecθ=2secθ
dx = asecthetatanthetaddx=asecθtanθdtheta = 2secthetatanthetadθ=2secθtanθdthetaθ
int x^2/(sqrt(x^2-4))dx = int (4sec^2theta)/(cancel(2tantheta))*cancel(2)secthetacancel(tantheta)dtheta
= int 4sec^3thetadtheta
= 4int sec^2thetasecthetadtheta
From here, we have a pretty difficult integral... We have to use a trick with integration by parts. Notice how this can be written as:
= 4int (1+tan^2theta)secthetadtheta
Let:
u = sectheta
v = tantheta
du = secthetatanthetadtheta
dv = sec^2thetadtheta
= 4[overbrace(secthetatantheta)^(u*v) - intoverbrace(tantheta)^(v)overbrace(secthetatanthetadɵ)^(du)]
= 4secthetatantheta - 4int(sec^2theta - 1)secthetadtheta
= 4secthetatantheta - 4intsec^3theta - secthetadtheta
= 4secthetatantheta - 4intsec^3thetadtheta + 4intsecthetadtheta
= 4secthetatantheta + 4ln|tantheta+sectheta| - 4intsec^3thetadtheta
where we already are supposed to know that intsecthetadtheta = ln|tantheta+sectheta|. You should have been shown this in class prior to doing this integral.
Now here's the weird part. The whole time we were solving intsec^3thetadtheta, so now we can explicitly say that and use that knowledge.
4int sec^3thetadtheta = 4secthetatantheta + 4ln|tantheta+sectheta| + C - 4intsec^3thetadtheta
8int sec^3thetadtheta = 4secthetatantheta + 4ln|tantheta+sectheta| + C
int sec^3thetadtheta = 1/2secthetatantheta + 1/2ln|tantheta+sectheta| + C
where C is now a new constant, 1/8C, which we don't need to know.
And finally, we can plug things back in.
sectheta = x/2
tantheta = sqrt(x^2-4)/2
We get:
= x/4(sqrt(x^2-4))/2 + 1/2ln|sqrt(x^2-4)/2+x/2| + C
