Let #u = x - 1# and #dv = sqrt(2 + x)dx#. By the product rule, #du = dx#. To integrate #dv# however, we will need to make a substitution.
Let #n = 2 + x#. Then #dn = dx#.
#dv = sqrt(n)dn#
#intdv = intsqrt(n)dn#
#v = 2/3n^(3/2) -># We can add #C# later on
#v = 2/3(2 + x)^(3/2)#
Now use the integration by parts formula.
#intudv = uv - intvdu#
#int(x - 1)sqrt(2 + x)dx = (x - 1)2/3(2 + x)^(3/2) - int(2/3(2 + x)^(3/2))dx#
You're going to need substitution to evaluate this integral. Let #m = 2 + x#. Then #dm = dx#.
#int2/3(2 + x)^(3/2)dx = int2/3m^(3/2)dm#
Now use #intx^ndx = x^(n + 1)/(n+1) + C#, as we did above.
#int2/3(2 + x)^(3/2)dx = 4/15m^(5/2)#
Reverse the substitution:
#int2/3(2 + x)^(3/2)dx = -4/15(2 + x)^(5/2)#
We can now put everything together.
#int(x - 1)sqrt(2 + x)dx = (x - 1)2/3(2 + x)^(3/2) + 4/15(2 + x)^(5/2) + C#
Hopefully this helps!
