Thread: Integration by substitution, what am I doing wrong

1. Integration by substitution, what am I doing wrong

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.

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

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

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

$\displaystyle dx = u du$

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

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

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

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

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

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

2. Originally Posted by 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
By factoring out $\displaystyle u^3$ in the last line we have

$\displaystyle \frac{1}{30} (u^3)(3u^2-5)$

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

$\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 3 (2x+1) - 5)$

$\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 6x+ 3-5)$

$\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 6x-2)$

$\displaystyle =\frac{1}{30} \sqrt{2x+1}^3 [2( 3x-1 ) ]$

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

3. Originally Posted by 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.

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

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

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

$\displaystyle dx = u du$

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

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

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

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

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

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
$\displaystyle \int{x\sqrt{2x + 1}\,dx}$

Let $\displaystyle u = 2x + 1$, then $\displaystyle x = \frac{u - 1}{2}$ and $\displaystyle \frac{du}{dx} = 2$.

Go from here. Remember that $\displaystyle \sqrt{u} = u^{\frac{1}{2}}$.