$$\int\limits_{}^{}2x(2x-1)^{1/2}dx$$
I know I need to solve by substitution so

u= 2x -1
du/dx = 2
how do I solve for x?

Question

$$\int\limits_{}^{}2x(2x-1)^{1/2}dx$$
I know I need to solve by substitution so

u= 2x -1
du/dx = 2
how do I solve for x?

anonymous · Answer

solve 
$$u=2x-1$$ for x as in elementary algebra

anonymous · Answer

$$x=\frac{u+1}{2}$$

anonymous · Answer

but why maybe I'm brain dead today Im just not getting it :L

turingtest · Answer

$$u=2x-1\implies 2x=u+1$$

amistre64 · Answer

no need to solve for "x"; just solve for "2x"

anonymous · Answer

good point!

anonymous · Answer

oh lol wow I'm dumb ha

anonymous · Answer

Or put x=2x and solve for z.

amistre64 · Answer

not that solveing for x is bad :)

anonymous · Answer

Thanks

anonymous · Answer

so 
u= 2x -1 
du/dx = 2
u = 2x - 1
(u + 1)/2 = x

$$\int\limits_{}^{} u^{1/2}(1 + u)/2$$
=

1/2 (u^(1/2) + u^(3/2))du
=
1/2(2u^(3/2)/3 + 2u^(5/2)/5) + c

anonymous · Answer

not subing but thanks for the help :)

turingtest · Answer

that is a sub
 now just sub back in 2x-1 for u and you have the final answer