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I am given the following indefinite integral:

$\int x\sqrt{1+2x}~dx$ using the substitution $u=\sqrt{1+2x}$

I am not sure where to begin, I know that $du=\dfrac{1}{\sqrt{1+2x}}$

But how can I get this integral in terms of $u$ when there is a $x$ in front?

Any help would be appreciated.


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We have $du=\frac{1}{\sqrt{1+2x}}\,dx$, so $dx=u\,du$. Note that $x=\frac{u^2-1}{2}$. So with the substitution you are using, we end up with $\int \frac{u^2-1}{2}u^2\,du$. Multiply through and integrate. – André Nicolas Feb 1 '14 at 1:49
I tried that But I am still getting the wrong answer. This is what I did $\frac{1}{2} \int u^4 - u^2 du$ Is this correct? Integrating gives me $\frac{u^5}{10} - \frac{u^3}{6} + C$ It seems to be still incorrect. – MathMajor Feb 1 '14 at 1:53
Yes, it is correct. – André Nicolas Feb 1 '14 at 1:54
So the result i get is: $1/10\, \left( 1+2\,x \right) ^{5/2}-1/6\, \left( 1+2\,x \right) ^{3/2}$ Which still seems to be incorrect... – MathMajor Feb 1 '14 at 1:59
Apart from the missing $+C$, it is correct. The expression can be manipulated into various equivalent forms. – André Nicolas Feb 1 '14 at 2:03
up vote 1 down vote accepted

Hint : Use $x=\frac{u^2-1}{2}$ to get dx = u du

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Try $u = 1 + 2x$, so $x = (u - 1)/2$. This will make the integral easy.

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Thanks for the suggestion but the problem requires that I use the given U substitution as noted above. – MathMajor Feb 1 '14 at 1:41
What you see above is more work, but it will do the job with that substitution. – ncmathsadist Feb 1 '14 at 1:44

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