Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 34

int_0^1r^3/sqrt(4+r^2)dr
Let's first evaluate the indefinite integral using the method of substitution,
Substitute x=4+r^2, =>r^2=x-4
=>dx=2rdr
intr^3/sqrt(4+r^2)dr=int(x-4)/(2sqrt(x))dx
=1/2int(x/sqrt(x)-4/sqrt(x))dx
=1/2int(sqrt(x)-4/sqrt(x))dx
=1/2((x^(1/2+1)/(1/2+1))-4(x^(-1/2+1)/(-1/2+1)))
=1/2((x^(3/2)/(3/2))-4(x^(1/2)/(1/2)))
=x^(3/2)/3-4x^(1/2)
substitute back x=r^2+4 and add constant C to the solution,
=(r^2+4)^(3/2)/3-4(r^2+4)^(1/2)+C
Now let's evaluate the definite integral,
int_0^1r^3/sqrt(4+r^2)dr=[(r^2+4)^(3/2)/3-4(r^2+4)^(1/2)]_0^1
=[(1^2+4)^(3/2)/3-4(1^2+4)^(1/2)]-[(0^2+4)^(3/2)/3-4(0^2+4)^(1/2)]
=[(5)^(3/2)/3-4(5)^(1/2)]-[4^(3/2)/3-4(4^(1/2)]
=[(5^(3/2)-12(5)^(1/2))/3]-[(2^2)^(3/2)/3-4*2]
=[5^(1/2)((5-12)/3)]-[2^3/3-8]
=[-7/3sqrt(5)]-[8/3-8]
=(-7/3sqrt(5))-((8-24)/3)
=-7/3sqrt(5)-(-16/3)
=-7/3sqrt(5)+16/3
=1/3(16-7sqrt(5))