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!
