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.
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.
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!