Integrate sqrt(x^2 + x) dx

Question

I have completed the squares and got x^2-x = (x+1/2)^2 -1/4

Now i don't know how to proceed, plz help me!

Answer 1

You have ∫√(u^2-a^2) du, so set

u = a secθ
du = a secθ tanθ
u^2-a^2 = a^2 tan^2θ

so it becomes

∫ (a tanθ)(a secθ tanθ) dθ
= a^2∫secθ tan^2θ dθ
= a^2∫sec^3θ - secθ dθ

Now, we all know that ∫secθ dθ = log(secθ + tanθ)

∫sec^3θ dθ can be done using integration by parts:
u = secθ, du = secθ tanθ dθ
dv = sec^2θ dθ, v = tanθ
so we have

∫ u dv = uv - ∫ v du
and then you have to go through it again, as is often the case with powers of trig functions.

Or, you can say that in the original integral,
u = a coshθ
du = a sinhθ dθ
u^2-a^2 = a^2 sinh^2θ
and the integrand becomes
(a sinhθ)(a sinhθ dθ)
= a^2 sinh^2 θ dθ
and you can again do that with integration by parts twice.