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**Tweety** Using $\displaystyle u = \sqrt{2x+1} $, show that

$\displaystyle \int x(\sqrt{2x+1} dx = \frac{1}{15}(2x+1)^{\frac{3}{2}} (3x-1) + c $

I have done this so many times and keep getting the wrong answer, heres my working.

$\displaystyle u = \sqrt{2x+1} $

$\displaystyle x = \frac{1}{2}(u^{2} -1) $

$\displaystyle \frac{dx}{du} = u $

$\displaystyle dx = u du $

$\displaystyle \int \frac{1}{2}(u^{2} -1) u u du $

$\displaystyle \frac{1}{2} \int (u^{4} -u^{2}) du $

$\displaystyle \frac{u^{5}}{5} - \frac{u^{3}}{3} $

$\displaystyle \frac{1}{2} (\frac{3u^{5} -5u^{3}}{15} ) $

$\displaystyle \frac{1}{2} \times \frac{1}{15} ( 3u^{5} - 5u^{3}) $

From here I am unable to get the desired expression, any help suggestions as to where I am going wrong would be greatly appreciated.

thanks