# Tricky integral using substitution

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• Feb 22nd 2011, 11:49 AM
isx99
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.
• Feb 22nd 2011, 12:02 PM
TheCoffeeMachine
The answers are equal.
• Feb 22nd 2011, 12:03 PM
pickslides
Try taking out $\displaystyle (2x-1)^{\frac{3}{2}}$ as a common factor.
• Feb 22nd 2011, 12:04 PM
pickslides
Quote:

Originally Posted by TheCoffeeMachine
The answers are equal.

Hey TCM, the OP has acknowledged this.
• Feb 22nd 2011, 12:05 PM
TheCoffeeMachine
Quote:

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

Oops, thanks! My apologies to the OP.
• Feb 22nd 2011, 12:13 PM
isx99
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:

$\frac{1}{2}(2x\ -1)^\frac{3}{2}(\frac{1}{5}(2x\ -1)\ + \frac{1}{3})$
• Feb 22nd 2011, 12:14 PM
isx99
Quote:

Originally Posted by TheCoffeeMachine
Oops, thanks! My apologies to the OP.

That's fine.
• Feb 22nd 2011, 12:21 PM
pickslides
Quote:

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
• Feb 22nd 2011, 12:38 PM
isx99
Thanks very much
Quote:

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

(Adding the fractions)
$-\frac{1}{10}\ +\ \frac{1}{6} = \frac{4}{60}\ =\ \frac{1}{15}$.

(Simplifying)
$(2x\ -1)^\frac{3}{2}(\frac{1}{5}x\ -\ \frac{1}{15})$.

(Taking out the $\frac{1}{15}$)
$\frac{1}{15}(2x\ -\ 1)^\frac{3}{2}(3x\ -\ 1)$, which is the answer given (Happy).