# Math Help - Tricky integral using substitution

1. ## Tricky integral using substitution

The Problem
Use the substitution $u = 2x - 1$ to find the integral of $f(x) = x\sqrt{2x - 1}$

Attempt
$\frac{du}{dx} = 2x$, so $dx = \frac{1}{2} du$.
Therefore, this integral simplifies to $\frac{1}{2}(Int\ x\sqrt{u}\ du)$.
Rearranging, we get x = $\frac{u + 1}{2}$, so this becomes $\frac{1}{2}$ x $\frac{1}{2}\ Int (\sqrt{u}\ (u+1))\ du$.
Expanding, this becomes $\frac{1}{4}\ Int\ (u^\frac{3}{2}\ + \sqrt{u})\ du$.
As everything is now a power of $u$, we can integrate as follows:
$\frac{1}{4}(\frac{u^\frac{5}{2}}{\frac{5}{2}}\ + \ \frac{u^\frac{3}{2}}{\frac{3}{2}})$.
Multiplying out the fractions:
$\frac{1}{4}(\frac{2u^\frac{5}{2}}{5}\ +\ \frac{2u^\frac{3}{2}}{3}})$
Multiplying out the brackets:
$\frac{2u^\frac{5}{2}}{20}\ +\ \frac{2u^\frac{3}{2}}{12}$
Cancelling:
$\frac{u^\frac{5}{2}}{10}\ + \ \frac{u^\frac{3}{2}}{6}$
And as $u = 2x - 1$, we get:
$\frac{1}{10}(2x\ -\ 1)^\frac{5}{2} + \frac{1}{6}(2x\ -\ 1)^\frac{3}{2}+ C$.

I've checked this using 'Show Steps' after inputting the integral into WolframAlpha, and it appears to be correct.

However, the answer given in both the textbook and WolframAlpha is:
$\frac{1}{15}(2x\ -\ 1)^\frac{3}{2}(3x\ +\ 1)+C$.

I would like a hint as to how I would progress from my answer to the answer given.

3. Try taking out $\displaystyle (2x-1)^{\frac{3}{2}}$ as a common factor.

4. Originally Posted by TheCoffeeMachine
Hey TCM, the OP has acknowledged this.

5. Originally Posted by pickslides
Hey TCM, the OP has acknowledged this.
Oops, thanks! My apologies to the OP.

6. ## OK, so what do I do next?

OK, so I took out the common factor to get:
$(2x-1)^\frac{3}{2}(\frac{1}{10}(2x\ - 1)\ + \frac{1}{6})$, then I could take out the factor of $\frac{1}{2}$ to get:

$\frac{1}{2}(2x\ -1)^\frac{3}{2}(\frac{1}{5}(2x\ -1)\ + \frac{1}{3})$

7. Originally Posted by TheCoffeeMachine
Oops, thanks! My apologies to the OP.
That's fine.

8. Originally Posted by isx99
OK, so I took out the common factor to get:
$(2x-1)^\frac{3}{2}(\frac{1}{10}(2x\ - 1)\ + \frac{1}{6})$, then I could take out the factor of $\frac{1}{2}$ to get:

$\frac{1}{2}(2x\ -1)^\frac{3}{2}(\frac{1}{5}(2x\ -1)\ + \frac{1}{3})$
Multiply the $\frac{1}{10}$ into the brackets, group like terms, simplitfy and then take back out $\frac{1}{15}$ you'll end up with the desired answer

9. ## Thanks very much

Originally Posted by pickslides
Multiply the $\frac{1}{10}$ into the brackets, group like terms, simplitfy and then take back out $\frac{1}{15}$ you'll end up with the desired answer
(That should be simplify, not simplitfy)

Thanks very much. Just to confirm, would this be:

(Expanding the brackets)
$(2x\ -1)^\frac{3}{2}(\frac{2x}{10}\ -\ \frac{1}{16}\ +\ \frac{1}{6}) \\ = \ (2x\ -1)^\frac{3}{2}(\frac{1}{5}x\ -\ \frac{1}{10}\ +\ \frac{1}{6})$.

$-\frac{1}{10}\ +\ \frac{1}{6} = \frac{4}{60}\ =\ \frac{1}{15}$.
$(2x\ -1)^\frac{3}{2}(\frac{1}{5}x\ -\ \frac{1}{15})$.
(Taking out the $\frac{1}{15}$)
$\frac{1}{15}(2x\ -\ 1)^\frac{3}{2}(3x\ -\ 1)$, which is the answer given .