The simplified answer is here.
Read below for the work.
Using the following identity:
tan^2x + 1 = sec^2x
the following substitution works out:
x = 2tantheta
dx = 2sec^2thetad theta
x^3 = 8tan^3theta
sqrt(x^2 + 4) = 2sectheta
Now we get:
int 8tan^3theta * 2sectheta * 2sec^2thetad theta
= 32int tan^3theta sec^3thetad theta
= 32int tan^2thetatantheta sec^3thetad theta
Now let's separate it so hopefully we get something we can do with u-substitution.
= 32int (sec^2theta - 1)tantheta sec^3thetad theta
= 32int sec^5thetatantheta - sec^3thetatanthetad theta
Notice how we can force this to look like u^4du - u^2du:
= 32int (sectheta)^4secthetatantheta - (sectheta)^2secthetatanthetad theta
Now, we can do another substitution:
u = sectheta
du = secthetatanthetad theta
This gives:
= 32int u^4 - u^2 du
= 32(u^5/5 - u^3/3)
= 32(sec^5theta/5 - sec^3theta/3)
Now, since 2sectheta = sqrt(x^2 + 4) -> sectheta = sqrt(x^2 + 4)/2:
= 32((x^2 + 4)^(5/2)/(2^5*5) - (x^2 + 4)^(3/2)/(2^3*3))
For convenience I've purposefully multiplied the second fraction by 4/4.
= cancel(32)((x^2 + 4)^(5/2)/(cancel(32)*5) - (4(x^2 + 4)^(3/2))/(cancel((4*8))*3))
= color(green)((x^2 + 4)^(5/2)/(5) - (4(x^2 + 4)^(3/2))/(3) + C)
If you want, you can stop here, but Wolfram Alpha gives you a different answer, so I'll show you how to get there too if you use that to check. Alternatively you could say "Is A = B" and it'll check for you, where A = one answer and B = the one you're checking.
Factor out 1/5:
= 1/5[(x^2 + 4)^(5/2) - (20(x^2 + 4)^(3/2))/(3)]
Factor out 1/3:
= 1/15[3(x^2 + 4)^(5/2) - 20(x^2 + 4)^(3/2)]
Factor out (x^2 + 4)^(3/2):
= 1/15(x^2 + 4)^(3/2)[3(x^2 + 4) - 20]
Distribute and add:
= 1/15(x^2 + 4)^(3/2)[3x^2 + 12 - 20]
= color(blue)(1/15(x^2 + 4)^(3/2)[3x^2 - 8] + C)
There it is.
