Integration by substitution, what am I doing wrong

Sep 2008
631
2
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
 

simplependulum

MHF Hall of Honor
Jan 2009
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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 \(\displaystyle u=\sqrt{2x+1}\)

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

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

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

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

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

MHF Helper
Aug 2008
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4,999
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 \(\displaystyle \sqrt{u} = u^{\frac{1}{2}}\).