Evaluate the following integral # x^3sqrt(1+x^2)#?

#int x^3 (1+x^2)^(1/2) dx#

Let #u = 1+x^2#.

This make #du = 2x dx# and #x^2 = u-1#

# = 1/2 int x^2 (1+x^2)^(1/2) 2x dx = 1/2 int(u-1)u^(1/2) du#

# = 1/2 int (u^(3/2)-u^(1/2)) du#

# = 1/2[2/5u^(5/2) - 2/3 u^(3/2)] +C#

# = 1/15 [3(1+x^2)^(5/2) - 5(1+x^2)^(3/2)] +C#

# = 1/15 (1+x^2)^(3/2)[3+3x^2-5 ] +C#

# = 1/15 (x^2-1)^(3/2) ( 3x^2-2) +C#

#int x^3 (1+x^2)^(1/2) dx#

We can integrate #t = (1+x^2)^(1/2)# by substitution if we think of one of the #x#'s as attached to #dx# to be #dt = 2x dx#.

#int x^2 (1+x^2)^(1/2) x dx#

Let #u = x^2# and #dv = (1+x^2)^(1/2) x dx#,

so #du = 2x dx# and #v = 1/3(1+x^2)^(3/2)#

#uv-int v du = 1/3 x^2 (1+x^2)^(3/2) - 2/3 int x (1+x^2)^(3/2) dx #

We can evaluate the integral by substitution, to get

# = 1/3 x^2 (1+x^2)^(3/2) - 2/3 1/5 (1+x^2)^(5/2) +C#

# = 1/15(x^2 +1)^(3/2)[5x^2-2(x^2+1)]#

# = 1/15(x^2 +1)^(3/2)(3x^2-2)#

Let me try another approach:

#d/(dx) x^n (1+x^2)^(3/2) = nx^(n-1) (1+x^2)^(3/2) + 3/2x^n(2x) (1+x^2)^(1/2)#

#color(white)(d/(dx) x^n (1+x^2)^(3/2)) = nx^(n-1) (1+x^2)^(3/2) + 3x^(n+1) (1+x^2)^(1/2)#

#color(white)(d/(dx) x^n (1+x^2)^(3/2)) = (nx^(n-1) (1+x^2)+3x^(n+1))(1+x^2)^(1/2)#

#color(white)(d/(dx) x^n (1+x^2)^(3/2)) = ((n+3)x^(n+1)+nx^(n-1))(1+x^2)^(1/2)#

In particular, putting #n=2# then #n=0# we find:

#d/(dx) (1/5x^2(1+x^2)^(3/2)) = (x^3+2/5x)(1+x^2)^(1/2)#

#d/(dx) (-2/15(1+x^2)^(3/2)) = -2/5x(1+x^2)^(1/2)#

Hence:

#int x^3(1+x^2)^(1/2) dx = (1/5x^2-2/15)(1+x^2)^(3/2) + C#

#color(white)(int x^3(1+x^2)^(1/2) dx) = 1/15(3x^2-2)(1+x^2)^(3/2) + C#