find the integral of?

Question

(2x-3)sqrt(x-2) or
(2x-3)(x-2)^(1/2)

Answer 1

Let u=inside of square root, so let u=(x-2) and then du=dx and 2x-3= 2(u+2)-3 = 2u+1 so then the integrand becomes

(2x-3)*sqrt(x-2)= (2u+1)*sqrt(u) = (2u+1)*u^1/2 then distribute to get 2u^(3/2)+1u^(1/2) then use the power rule and finally end up with 4/5*u^(5/2)+2/3*u^(3/2)+C and finally plug back in u=x-2

Answer 2

So you want to solve

Integral [(2x - 3) (x - 2)^(1/2)]dx

There are many methods of doing this, but one thing you can use is u substitution in a slightly different manner than usual.

Let u = x - 2.

What we want to do is manipulate this expression into what it is being multiplied to. We want to try and get 2x - 3 from that, and we can.

u + 2 = x
2(u + 2) = 2x
2u + 4 = 2x
2u + 1 = 2x - 3

Therefore
2du = 2dx, and
du = dx

Integral [(2x - 3) (x - 2)^(1/2)]dx, after u substitution, becomes

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

Note that at this point, we can actually multiply the monomial u^(1/2) with the binomial. Remember that 2u times u^(1/2) means we can add the exponents, to get 2u^(3/2).

Integral [2u^(3/2) + u^(1/2)]du

Now, we can split this integral of a sum to the sum of two integrals.

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

And we can pull out all constants; in this case there's just the 2.

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

And now we easily integrate this using the reverse power rule.

2 * (2/5)(u^(5/2)) + (2/3)u^(3/2) + C

Combining the constants,

(4/5)(u^(5/2)) + (2/3)u^(3/2) + C

And now, since we let u = x - 2, our final answer is

(4/5)( [x - 2]^(5/2) ) + (2/3) ( [x - 2]^(3/2) ) + C

We can simplify this even more by factoring, but this should be a good enough answer when it's time for the final exam.

Answer 3

Let
u = x - 2
x = u + 2
du = dx

∫(2x-3)sqrt(x-2)dx = ∫(2u+4-3)sqrt(u)du = ∫(2u+1)sqrt(u)du
= ∫[(2u^(3/2) + u^(1/2)]du = (4/5)u^(5/2) + (2/3)u^(3/2) + C
= (4/5)(x - 2)^(5/2) + (2/3)(x - 2)^(3/2) + C

Answer 4

use parts, it will make your life easier.... equation is vu-Svdu
let u = (2x-3), du = 2, dv = (x-2)^(1/2), v = (2/3)(x-2)^(3/2),
ie, (2x-3)(2/3)(x-2)^(3/2) - S 2(2/3)(x-2)^(3/2)dx
(2x-3)(2/3)(x-2)^(3/2) - (4/15)(x-2)^(5/2)

Answer 5

[2xsqrt(-2x)(2x-5)]/5