I'm struggling solving the following indefinite integral with u-substition.
Given:
$\int 2\sqrt[]{2x-1} dx , u=2x-1$
I begin to solve by factoring out 2, thus:
$2 \int \sqrt[]{2x-1} dx$
and rewrite with an exponent and u-substitution:
$2 \int u^{1/2} du$
which evaluates to: $\frac{2}{3}u^{3/2} + c$
"Unsubstituting" u gives:
$\frac{2}{3}(2x-1)^{3/2} + c$ (which is the correct answer on my homework sheet)
but what about the 2 we factored out in the beginning? Is not the final solution:
$2\cdot\frac{2}{3}(2x-1)^{3/2} + c$
thus:
$\frac{4}{3}(2x-1)^{3/2} + C$
$\endgroup$2 Answers
$\begingroup$You have performed all the steps correctly. There is just one minor error.
In u-substitution, you set your u to be:
$$u = 2x -1$$
Therefore differentiating, we get:
$$du = 2dx,\ dx = \frac{du}{2}$$
When you substitute your $u$ into the integral you also have to make sure to substitute your $du$, or else you're taking the integral with respect with another variable.
$\endgroup$ 1 $\begingroup$Your homework sheet appears to be correct. You forgot to account for the change from $dx$ to $du$. If $u=2x-1$, then $du=2dx$. When you substitute, you should have
$$\int2\sqrt{2x-1}dx=\int\sqrt{2x-1}(2dx)=\int\sqrt udu$$
$\endgroup$ 3
