Can someone help me integrate using substitution?

Question

Can someone show me (step by step) how to integrate x(2x-1)^(1/2)dx? I have the answer, but I'm not sure how to work the problem.

Answer 1

So you want to solve

Integral ( x(2x - 1)^(1/2) ) dx

You're going to solve this using substitution, in a slightly different way.

Let u = 2x - 1. At this point, rather than solving for du directly, you're going to first solve for x, and then solve for dx.

u + 1 = 2x
(1/2) (u + 1) = x

Differentiating both sides,

(1/2)du = dx

Now, we make the appropriate substitution, Remember we want to substitute the x, the (2x - 1), and the dx, and we have all these values.

Integral ( x(2x - 1)^(1/2) ) dx becomes

Integral ( [(1/2)(u + 1)] [u^(1/2)] [(1/2)du] )

Let's pull out ALL constants from the integral. This is a step that makes integration easier, and it's a perfectly valid step too. Our constants in this case are (1/2) and another (1/2), so pulled out together, they become 1/4, and the integral becomes

(1/4) * Integral ( (u + 1) u^(1/2) du)

Note that we can now distribute the u^(1/2). Note that u * u^(1/2) is u^(3/2), so we have

(1/4) * Integral (u^(3/2) + u^(1/2)) du

Now, we integrate normally, using the reverse power rule.

(1/4) * [ (2/5)u^(5/2) + (2/3)u^(3/2) ] + C

Let's clean this up by multiplying the fractions out.

(1/10)u^(5/2) + (1/6)u^(3/2) + C

And now, we substitute in u = (2x - 1) back, to get

(1/10)(2x - 1)^(5/2) + (1/6)(2x - 1)^(3/2) + C

Answer 2

So Y=∫x*sqrt(2x-1)*dx,
assume z= sqrt(2x-1) or zz= 2x-1 or x=0.5(zz+1) and dx = zdz;
Y=∫(0.5(zz+1)) *z*(zdz) =
= 0.5∫(zzzz +zz)dz = 0.5((1/5)z^5 +(1/3)z^3) =
= (1/30)(3zz +5)zzz =
= (1/30)(3(2x-1) +5)* (2x-1)^(3/2) =
=(1/15) (3x+1) (2x-1)^(3/2) +C;

puggy's right!