1. ## Another integration problem

How do I calculate the integral when the denominator is a square root of a function?

e.g.

$\displaystyle \int \frac{u^3+1}{\sqrt{u^4+4u}} du$

2. This one you can solve using a substitution.

$\displaystyle \int{\frac{u^3 + 1}{\sqrt{u^4 + 4u}}\,du} = \frac{1}{4}\int{(u^4 + 4u)^{-\frac{1}{2}}(4u^3 + 4)\,du}$.

Now let $\displaystyle v = u^4 + 4u$ so that $\displaystyle dv = (4u^3 + 4)\,du$ and the integral becomes

$\displaystyle \frac{1}{4}\int{v^{-\frac{1}{2}}\,dv}$

$\displaystyle = \frac{1}{4}\left(\frac{v^{\frac{1}{2}}}{\frac{1}{2 }}\right) + C$

$\displaystyle = \frac{1}{4}\left(2\sqrt{v}\right) + C$

$\displaystyle = \frac{\sqrt{u^4 + 4u}}{2} + C$.

3. Originally Posted by blackdragon190
How do I calculate the integral when the denominator is a square root of a function?
$\displaystyle \int \frac{u^3+1}{\sqrt{u^4+4u}} du$
What is the derivative of $\displaystyle \dfrac{\sqrt{u^4+4u}}{2}?$

$\displaystyle \displaystyle \frac{u^{3}+1}{\sqrt{u^{4}+4\ u}} = \frac{1}{4}\ \frac{4\ u^{3}+4}{\sqrt{u^{4}+4\ u}}$ (1)

... and that...

$\displaystyle \displaystyle \frac{d}{d u}\ (u^{4}+4\ u) = 4\ u^{3} + 4$ (2)

At this point You are 'almost arrived' ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. The standard first step is to rationalize the denominator by multiplying the denominator and numerator by the expression sitting in the denominator.

6. Originally Posted by wonderboy1953
The standard first step is to rationalize the denominator by multiplying the denominator and numerator by the expression sitting in the denominator.
Sorry, but this is not correct.

7. Thanks! I was just forgetting that a square root in the denominator means that it's to the power of -1/2.