I'm having trouble figuring out the substitution for $\displaystyle \int \frac {5-2\sqrt{x}}{x^3}\frac d{dx}$.

I tried $\displaystyle u=x^3$ and $\displaystyle u=5-2\sqrt{x}$, but those don't seem to lead me anywhere.

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- Oct 4th 2008, 01:33 PMdiablo2121Help with indefinite integral substitution
I'm having trouble figuring out the substitution for $\displaystyle \int \frac {5-2\sqrt{x}}{x^3}\frac d{dx}$.

I tried $\displaystyle u=x^3$ and $\displaystyle u=5-2\sqrt{x}$, but those don't seem to lead me anywhere. - Oct 4th 2008, 01:41 PMicemanfan
I assume you mean this:

$\displaystyle \int \frac{5 - 2\sqrt{x}}{x^3} \, dx$

Use the old single term in the denominator trick.

$\displaystyle \int \frac{5}{x^3} - \frac{2}{x^{5/2}} \, dx$ is a lot easier to work with. - Oct 4th 2008, 02:41 PMdiablo2121
What about $\displaystyle \int \frac{x}{x^4+1} \, dx$?

- Oct 4th 2008, 08:24 PMFire Mage
This fits a pattern. Look at it as:

$\displaystyle \int\frac{du}{(a^2 + u^2)}$

That fits the pattern for arctan, so:

u = x^2

du = 2xdx

a = 1

You have to accomodate for the 2 in 2xdx, so:

$\displaystyle \frac{1}{2}\int\frac{2xdx}{1+x^4}$

therefore, the basic integral is:

$\displaystyle \frac{1}{a}\arctan\frac{u}{a}+C$

So from there on it is plugging in numbers! You have to learn these patterns so read up!