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?

Answer 1

solve
\[u=2x-1\] for x as in elementary algebra

Answer 2

\[x=\frac{u+1}{2}\]

Answer 3

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

Answer 4

\[u=2x-1\implies 2x=u+1\]

Answer 5

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

Answer 6

good point!

Answer 7

oh lol wow I'm dumb ha

Answer 8

Or put x=2x and solve for z.

Answer 9

not that solveing for x is bad :)

Answer 10

Thanks

Answer 11

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

Answer 12

not subing but thanks for the help :)

Answer 13

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